I INDIA INTERNATIONAL SCHOOL
Class XI
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CONIC SECTIONS Ch-11
Sections of a cone
Let l be a fixed
vertical line and m
be another line
intersecting it at a
fixed point V and
inclined to it at an
angle α (Fig. 11.1).
Suppose we rotate
the line m around
the line l in such a
way that the angle
α remains
constant. Then the
surface generated is a double-napped right circular hollow cone
herein after referred as cone and extending indefinitely in both directions (Fig. 11.2). The
point V is called the vertex; the line l is the axis of the cone. The rotating line m is
called a generator of the cone. The vertex separates the cone into two parts called
nappes.
If we take the intersection of a plane with a cone, the section so obtained is called a
conic section. Thus, conic sections are the curves obtained by intersecting a right
circular cone by a plane.
We obtain different kinds of conic sections depending on the position of the intersecting
plane with respect to the cone and the angle made by it with the vertical axis of the
cone. Let β be the angle made by the intersecting plane with the vertical axis of the
cone (Fig.11.3).
The intersection of the plane with the cone can take place either at the vertex of the
cone or at any other part of the nappe either below or above the vertex.
When the plane cuts the nappe (other than the vertex) of the cone, we have the
following situations:
(a) When β = 90o, the section is a circle.
(b) When α < β < 90o, the section is an ellipse.
(c) When β = α; the section is a parabola.
(In each of the above three situations, the plane cuts entirely across one nappe
of the cone).
(d) When 0 ≤ β < α; the plane cuts through both the nappes and the curves of
intersection is a hyperbola.
Indeed, these curves are important tools for present day exploration of outer space and
also for research into the behaviour of atomic particles.
We take conic sections as plane curves. For this purpose, it is convenient to use equivalent
definition that refer only to the plane in which the curve lies, and refer to special points
and lines in this plane called foci and directrices. According to this approach, parabola,
ellipse and hyperbola are defined in terms of a fixed point (called focus) and fixed line
(called directrix) in the plane.
If S is the focus and l is the directrix, then the set of all points in the plane whose
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distance from S bears a constant ratio e called eccentricity to their distance from l is a
conic section.
As special case of ellipse, we obtain circle for which e = 0 and hence we study it
differently.
Circle A circle is the set of all points in a plane which are at a fixed distance
from a fixed point in the plane. The fixed point is called the centre of the circle and the
distance from centre to any point on the circle is called the radius of the circle.
1. Equation of a Circle in Standard Form
Let O (0, 0) be the centre of the circle and r be its radius. Let P
(x, y) be a point in the plane, then P lies on the circle if OP = r
i.e. √(𝑥 − 0)2 + (𝑦 − 0)2 = 𝑟
⇒ x2 + y2 = r2
This is standard form of the equation of a circle
2. Equation of a Circle in Central Form
Let C (h, k) be the centre of the circle and r be its radius. Let P
(x, y) be a point in the plane, then P lies on the circle if CP = r
⇒ √(𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟
⇒ (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2
This is known as the central form of the equation of a circle.
3. Equation of a Circle in Diameter Form
Let 𝐴(𝑥1 , 𝑦1 ) and 𝐵(𝑥2 , 𝑦2 ) be the extremities of a
diameter of the circle.
Let 𝑃(𝑥, 𝑦), be any point on the circle. Thus
𝑦−𝑦
slope of the line AP = 𝑥−𝑥1 and
1
𝑦−𝑦
slope of the line BP = 𝑥−𝑥2
2
Now P lies on the circle, so ∠APB = 90∘
The lines i.e. AP and BP are perpendicular to each
𝑦−𝑦1 𝑦−𝑦2
other. So,
⋅ 𝑥−𝑥 = −1
𝑥−𝑥
1
2
⇒ (y − y1 )(y − y2 ) = −(x − x1 )(x − x2 )
⇒ (x − x1 )(x − x2 ) + (y − y1 )(y − y2 ) = 0
This is the equation of a circle in diameter form.
4. Equation of a circle in General Form
We know that the equation of the circle with centre (ℎ, 𝑘) and radius 𝑟 is
(𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2
⇒ 𝑥 2 + 𝑦 2 − 2ℎ𝑥 − 2𝑘𝑦 + ℎ2 + 𝑘 2 − 𝑟 2 = 0
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It can be written as 𝑥 2 + 𝑦 2 + 2𝑔𝑥 + 2𝑓𝑦 + 𝑐 = 0
where 𝑔 = −ℎ, 𝑓 = −𝑘 and 𝑐 = ℎ2 + 𝑘 2 − 𝑟 2
∴ 𝑟 2 = ℎ2 + 𝑘 2 − 𝑐= (−𝑔)2 + (−𝑓)2 − 𝑐 = 𝑔2 + 𝑓 2 − 𝑐
But 𝑟 2 > 0, so 𝑔2 + 𝑓 2 − 𝑐 > 0
Now if we consider any equation 𝑥 2 + 𝑦 2 + 2𝑔𝑥 + 2𝑓𝑦 + 𝑐 = 0
Properties of Circles
1 If 𝑔2 + 𝑓 2 − 𝑐 = 0, the equation of circle is satisfied by one and only one point
(−𝑔, −𝑓). Therefore, it represents a single point known as point circle.
2 If 𝑔2 + 𝑓 2 − 𝑐 < 0, the equation of the circle is not satisfied by any real value of 𝑥, 𝑦
i.e. it is not satisfied by the co-ordinates of any point in the plane.
3 The general equation of a circle has the following characteristics.
(i) It is an equation of second degree in 𝑥, 𝑦 containing no product term 𝑥𝑦.
(ii) Coefficients of 𝑥 2 = Coefficients of 𝑦 2 = 1
4 The equation 𝑎𝑥 2 + 𝑎𝑦 2 + 2𝑔𝑥 + 2𝑓𝑦 + 𝑐 = 0 represents 𝑎 circle if 𝑎 ≠ 0 and
𝑔
𝑓
𝑔2 + 𝑓 2 − 𝑐 > 0. Its centre is (− 𝑎 , − 𝑎) and radius is
√𝑔2 +𝑓 2 −𝑎𝑐
|𝑎|
.
The equation 𝑎𝑥 2 + 2ℎ𝑥𝑦 + 𝑏𝑦 2 + 2𝑔𝑥 + 2𝑓𝑦 + 𝑐 = 0 represents a circle if
(𝑖) 𝑎 = 𝑏 ≠ 0
(ii) h = 0 and
(iii) 𝑔2 + 𝑓 2 − 𝑎𝑐 > 0
Note: The general equation of the circle involves three constants which implies that at least
three conditions are required to determine a circle uniquely.
Example 1 Find an equation of the circle with centre at (0,0) and radius 𝑟.
Solution Here ℎ = 𝑘 = 0. Therefore, the equation of the circle is 𝑥 2 + 𝑦 2 = 𝑟 2 .
Example 2 Find the equation of the circle with centre (−3,2) and radius 4 .
Solution Here ℎ = −3, 𝑘 = 2 and 𝑟 = 4. Therefore, the equation of the required circle is
(𝑥 + 3)2 + (𝑦 − 2)2 = 16
Example 3 Find the centre and the radius of the circle 𝑥 2 + 𝑦 2 + 8𝑥 + 10𝑦 − 8 = 0
Solution The given equation is
(𝑥 2 + 8𝑥) + (𝑦 2 + 10𝑦) = 8
Now, completing the squares within the parenthesis, we get
i.e. (𝑥 2 + 8𝑥 + 16) + (𝑦 2 + 10𝑦 + 25) = 8 + 16 + 25
(𝑥 + 4)2 + (𝑦 + 5)2 = 49
{𝑥 − (−4)}2 + {𝑦 − (−5)}2 = 72
Therefore, the given circle has centre at (−4, −5) and radius 7 .
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Example 4 Find the equation of the circle which passes through the points (2, −2), and (3,4)
and whose centre lies on the line 𝑥 + 𝑦 = 2.
Solution Let the equation of the circle be (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2.
Since the circle passes through (2, −2) and (3,4), we have
(2 − ℎ)2 + (−2 − 𝑘)2 = 𝑟 2
and (3 − ℎ)2 + (4 − 𝑘)2 = 𝑟 2
Also since the centre lies on the line 𝑥 + 𝑦 = 2, we have
ℎ+𝑘 =2
Solving the equations (1), (2) and (3), we get
ℎ = 0.7, 𝑘 = 1.3 and 𝑟 2 = 12.58
Hence, the equation of the required circle is (𝑥 − 0.7)2 + (𝑦 − 1.3)2 = 12.58.
EXERCISE 10.1
In each of the following Exercises 1 to 5, find the equation of the circle with
1
Centre (0,2) and radius 2.
Sol. Here ℎ = 0, 𝑘 = 2 and 𝑟 = 2.
Equation of circle is (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2
i.e., (𝑥 − 0)2 + (𝑦 − 2)2 = 22
or 𝑥 2 + 𝑦 2 − 4𝑦 + 4 = 4
or 𝑥 2 + 𝑦 2 − 4𝑦 = 0.
2
Centre (−2,3) and radius 4 .
Sol. Here ℎ = −2, 𝑘 = 3 and 𝑟 = 4.
Equation of circle is (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2
i.e., (𝑥 + 2)2 + (𝑦 − 3)2 = 42
Or 𝑥 2 + 4𝑥 + 4 + 𝑦 2 − 6𝑦 + 9 = 16
Or 𝑥 2 + 𝑦 2 + 4𝑥 − 6𝑦 − 3 = 0
3
1 1
1
Centre (2 , 4) and radius 12.
Sol. We know that the equation of the circle with centre at (ℎ, 𝑘) and radius 𝑟 is given by
(𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2 … (𝑖)
Here
1 1
1
(ℎ, 𝑘) = (2 , 4) and 𝑟 = 12
Putting values of ℎ, 𝑘 and 𝑟 in (𝑖),
1 2
1 2
1 2
∴ Equation of the required circle is (𝑥 − 2) + (𝑦 − 4) = (12)
Or
1
1
1
1
𝑥 2 + 4 − 𝑥 + 𝑦 2 + 16 − 2 𝑦 = 144
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1
11
Or 𝑥 2 + 𝑦 2 − 𝑥 − 2 𝑦 + 36 = 0
Multiplying by 36, we have
4
1
1
1
[∵ 4 + 16 − 144 =
36+9−1
144
11
= 36]
36𝑥 2 + 36𝑦 2 − 36𝑥 − 18𝑦 + 11 = 0.
Centre (1,1) and radius √2.
Sol. Here ℎ = 1, 𝑘 = 1 and 𝑟 = √2.
Equation of circle is (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2 i.e.,
(𝑥 − 1)2 + (𝑦 − 1)2 = (√2)2
𝑥 2 − 2𝑥 + 1 + 𝑦 2 − 2𝑦 + 1 = 2
Or
Or 𝑥 2 + 𝑦 2 − 2𝑥 − 2𝑦 = 0
5
Centre (−𝑎, −𝑏) and radius √𝑎2 − 𝑏 2 .
Sol. Here ℎ = −𝑎, 𝑘 = −𝑏 and 𝑟 = √𝑎2 − 𝑏 2 .
Equation of circle is (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2
i.e., (𝑥 + 𝑎)2 + (𝑦 + 𝑏)2 = 𝑎2 − 𝑏 2
or 𝑥 2 + 2𝑎𝑥 + 𝑎2 + 𝑦 2 + 2𝑏𝑦 + 𝑏 2 = 𝑎2 − 𝑏 2
Or
𝑥 2 + 𝑦 2 + 2𝑎𝑥 + 2𝑏𝑦 + 2𝑏 2 = 0
In each of the following Exercises 6 to 9, find the centre and radius of the circles.
6. (𝑥 + 5)2 + (𝑦 − 3)2 = 36.
Sol. The given equation is [𝑥 − (−5)]2 + (𝑦 − 3)2 = 62 .
Comparing it with (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2 ,
We have
ℎ = −5, 𝑘 = 3 and 𝑟 = 6.
∴ The given circle has centre at (ℎ, 𝑘) = (−5,3) and radius 𝑟 = 6.
7
𝑥 2 + 𝑦 2 − 4𝑥 − 8𝑦 − 45 = 0.
Sol. The given equation is
(𝑥 2 − 4𝑥) + (𝑦 2 − 8𝑦) = 45.
Completing the squares within the parenthesis, we get
1
1
[Adding ( 2 coeff. of 𝑥)2 and ( 2 coeff. of 𝑦)2 to both sides]
(𝑥 2 − 4𝑥 + 22 ) + (𝑦 2 − 8𝑦 + 42 ) = 45 + 4 + 16
⇒ (𝑥 − 2)2 + (𝑦 − 4)2
Comparing it with (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2, we have
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ℎ = 2, 𝑘 = 4 and 𝑟 = √65.
∴ The given circle has centre at (ℎ, 𝑘) = (2,4) and radius 𝑟 = √65.
8
𝑥 2 + 𝑦 2 − 8𝑥 + 10𝑦 − 12 = 0.
Sol. The given equation is (𝑥 2 − 8𝑥) + (𝑦 2 + 10𝑦) = 12.
Completing the squares within the parenthesis, we get
1
1
[Adding ( 2 coeff. of 𝑥)2 and ( 2 coeff. of 𝑦)2 to both sides]
(𝑥 2 − 8𝑥 + 42 ) + (𝑦 2 + 10𝑦 + 52 ) = 12 + 16 + 25
⇒ (𝑥 − 4)2 + (𝑦 + 5)2 = 53
⇒ (𝑥 − 4)2 + [𝑦 − (−5)]2 = (√53)2
Comparing it with (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2, we have
ℎ = 4, 𝑘 = −5 and 𝑟 = √53.
∴ The given circle has centre at (ℎ, 𝑘) = (4, −5) and radius 𝑟 = √53.
9. 2𝑥 2 + 2𝑦 2 − 𝑥 = 0.
Sol. The given equation is
2𝑥 2 + 2𝑦 2 − 𝑥 = 0
Dividing every term by 2 , to make coefficient of 𝑥 2 and 𝑦 2 unity,
1
or 𝑥 2 + 𝑦 2 − 2 𝑥 = 0
1
or (𝑥 2 − 2 𝑥) + 𝑦 2 = 0
Completing the square within parenthesis,
1
2
1 2
1 2
1 2
1
Adding (2 co-eff. of 𝑥) = (4) = 16 to both sides,
Or
(𝑥 − 4) + 𝑦 2 = (4)
1
1
Comparing it with (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2, we have ℎ = 4 , 𝑘 = 0 and 𝑟 = 4
1
1
∴ The centre of the given circle is (ℎ, 𝑘) = (4 , 0) and radius is 𝑟 = 4.
10. Find the equation of the circle passing through the points (4,1) and (6,5) and whose centre
is on the line 4𝑥 + 𝑦 = 16.
Sol. Let the equation of the circle be
(𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2
(𝑖)
Since the points (4,1) and (6,5) lie on it, we have
(4 − ℎ)2 + (1 − 𝑘)2 = 𝑟 2
(ii)
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And (6 − ℎ)2 + (5 − 𝑘)2 = 𝑟 2 … (iii)
From (ii) and (iii) equating the two values of 𝑟 2 , we have
(4 − ℎ)2 + (1 − 𝑘)2 = (6 − ℎ)2 + (5 − 𝑘)2
or (16 − 8ℎ + ℎ2 ) + (1 − 2𝑘 + 𝑘 2 ) = (36 − 12ℎ + ℎ2 ) +(25 − 10𝑘 + 𝑘 2 )
or − 8ℎ − 2𝑘 + 17 = −12ℎ − 10𝑘 + 61 ⇒ 4ℎ + 8𝑘 = 44
Dividing by 4,
… (iv)
ℎ + 2𝑘 = 11
Since the centre (ℎ, 𝑘) lies on the line 4𝑥 + 𝑦 = 16,
we have
4ℎ + 𝑘 = 16
… (v)
Let us solve eqns. (iv) and (𝑣) for ℎ and 𝑘.
Eqn. (v) −4 × eqn. (iv) gives 𝑘 − 8𝑘 = 16 − 44 or −7𝑘 = −28 or 𝑘 = 4
Putting 𝑘 = 4 in eqn. (iv), ℎ + 8 = 11 or ℎ = 3
Putting these values of ℎ and 𝑘 in (ii), we have
𝑟 2 = (4 − 3)2 + (1 − 4)2 = 1 + 9 = 10
Putting these values of ℎ, 𝑘 and 𝑟 2 in (i), the equation of required circle is
(𝑥 − 3)2 + (𝑦 − 4)2 = 10
Or 𝑥 2 + 9 − 6𝑥 + 𝑦 2 + 16 − 8𝑦 = 10
Or
𝑥 2 + 𝑦 2 − 6𝑥 − 8𝑦 + 15 = 0
11 Find the equation of the circle passing through the points (2,3) and (−1,1) and whose
centre is on the line 𝑥 − 3𝑦 − 11 = 0.
Sol. Let the equation of the circle be (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2 .. (𝑖)
Since the points (2,3) and (−1,1) lie on it, we have
(2 − ℎ)2 + (3 − 𝑘)2 = 𝑟 2
And
(−1 − ℎ)2 + (1 − 𝑘)2 = 𝑟 2
… (𝑖𝑖)
… (iii)
From (ii) and (iii), equating the two values of 𝑟 2 , we have
(2 − ℎ)2 + (3 − 𝑘)2 = (−1 − ℎ)2 + (1 − 𝑘)2
or (4 − 4ℎ + ℎ2 ) + (9 − 6𝑘 + 𝑘 2 ) = (1 + 2ℎ + ℎ2 ) + (1 − 2𝑘 + 𝑘 2 )
or 13 − 4ℎ − 6𝑘 = 2 + 2ℎ − 2𝑘
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… (iv)
or 6ℎ + 4𝑘 = 11
Given: The centre (ℎ, 𝑘) lies on the line 𝑥 − 3𝑦 − 11 = 0
∴ ℎ − 3𝑘 = 11
… (v)
Eqn. (iv) −6 × eqn. (v) gives 4𝑘 + 18𝑘 = 11 − 66
⇒ 22𝑘 = −55 ⇒ 𝑘 =
−55
5
=−
22
2
5
Putting 𝑘 = − 2 in (𝑣), we have
5
15
7
ℎ − 3 (− 2) = 11 or ℎ = 11 − 2 = 2
Putting the values of ℎ and 𝑘 in (ii), we get
7 2
5 2
(2 − ) + (3 + ) = 𝑟 2
2
2
4−7 2
6+5 2
9 121 130
(
) +(
) = 𝑟2 ⇒ 𝑟2 = +
=
2
2
4
4
4
Putting values of ℎ, 𝑘 and 𝑟 2 in (i),
7 2
5 2
130
required equation of circle is (𝑥 − 2) + (𝑦 + 2) = 4
49
25
130
Or (𝑥 2 − 7𝑥 + 4 ) + (𝑦 2 + 5𝑦 + 4 ) = 4
or 𝑥 2 + 𝑦 2 − 7𝑥 + 5𝑦 − 14 = 0
49
25
130
(∴ 4 + 4 − 4 =
74−130
4
=
−56
4
= −14)
12 Find the equation of the circle with radius 5 whose centre lies on 𝑥-axis and passes
through the point (2,3).
Sol. Let the centre of the circle be C(ℎ, 0)
[∵ Every point on 𝑥-axis has its 𝑦-coordinate 0]
Radius of circle = 5
∴ Equation of circle is (𝑥 − ℎ)2 + (𝑦 − 0)2 = 52
or 𝑥 2 + 𝑦 2 − 2ℎ𝑥 + ℎ2 − 25 = 0
Since it passes through the point (2,3), we have by putting
𝑥 = 2, 𝑦 = 3
Or 4 + 9 − 4ℎ + ℎ2 − 25 = 0
Or
ℎ2 − 4ℎ − 12 = 0
Or (ℎ − 6)(ℎ + 2) = 0
Or ℎ = 6 or h = −2
When ℎ = 6, from (𝑖), the equation of circle is
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𝑥 2 + 𝑦 2 − 12𝑥 + 11 = 0.
(∵ 36 − 25 = 11)
When ℎ = −2, from (i) the equation of circle is
𝑥 2 + 𝑦 2 + 4𝑥 − 21 = 0.
(∵ 4 − 25 = −21)
13 Find the equation of the circle passing through (0,0) and making intercepts 𝑎 and 𝑏 on the
coordinate axes.
Sol. Given: The required circle passes through the origin
O(0,0) and makes intercepts 𝑎 and 𝑏 on the co-ordinates
axes, therefore the circle also passes through the points
A(𝑎, 0) and B(0, 𝑏). Join AB. Let C be the centre of required
circle.
We know that angle between the co-ordinate axes i.e.,
∠AOB = 90∘ .
∴ AB is a diameter of the circle.
[Angle in a semi-circle]
∴ Centre C of the circle is the mid-point of diameter AB.
𝑎+0 0+𝑏
𝑎 𝑏
∴ By mid-point formula, Centre C is ( 2 , 2 ) = (2 , 2).
1
Also radius 𝑟 of the circle = AC = BC = 2 ( diameter AB)
=
1
1
√(𝑎 − 0)2 + (𝟎 − 𝑏)2 = √(𝑎2 + 𝑏)2
2
2
∴ Equation of required circle is(𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2 ,
𝑎 2
𝑏 2
𝑎 2
𝑏 2
1
i.e., (𝑥 − 2) + (𝑦 − 2 ) = (2 √𝑎2 + 𝑏 2 )
2
1
i.e., (𝑥 − 2) + (𝑦 − 2 ) = 4 (𝑎2 + 𝑏 2 )
𝑎2
𝑏2
𝑎2
𝑏2
or 𝑥 2 + 4 − 𝑎𝑥 + 𝑦 2 + 4 − 𝑏𝑦 = 4 + 4
or 𝑥 2 + 𝑦 2 − 𝑎𝑥 − 𝑏𝑦 = 0.
14 Find the equation of a circle with centre (2,2) and passing through the point (4,5).
Sol. Centre of circle is C(2,2).
Since the circle passes through the point P(4,5), its radius
𝑟 = 𝐶𝑃 = √(4 − 2)2 + (5 − 2)2 = √4 + 9 = √13
∴ The equation of circle is (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2
(Distance Formula)
(𝑥 − 2)2 + (𝑦 − 2)2 = (√13)2
Or (𝑥 2 − 4𝑥 + 4) + (𝑦 2 − 4𝑦 + 4) = 13
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Or 𝑥 2 + 𝑦 2 − 4𝑥 − 4𝑦 = 5
15 Does the point (−2.5,3.5) lie inside, outside or on the circle 𝑥 2 + 𝑦 2 = 25?
Sol. The given circle 𝑥 2 + 𝑦 2 = 25
or 𝑥 2 + 𝑦 2 = 52
has its centre at O(0,0) and radius 5 .
Distance of the given point P(−2.5,3.5) from centre 𝑂(0,0) is
OP = √(2.5)2 + (3.5)2 = √6.25 + 12.25
= √18.50 < √25 = 5 = radius
Since, OP is less than the radius of the circle, the given point lies inside the circle
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Parabola
A parabola is the set of points P whose distances from a fixed
point F in the plane are equal to their distances from a fixed
line l in the plane. The fixed point F is called focus and the
fixed line l the directrix of the parabola.
Standard equations of parabola
The four possible forms of parabola are shown below in Fig.
11.7 (a) to (d) The latus rectum of a parabola is a line segment
perpendicular to the axis of the parabola, through the focus and
whose end points lie on the parabola (Fig. 11.7).
Latus Rectum: Latus rectum of a parabola is a line segment perpendicular to the axis of
the parabola, through the focus and whose end points lie on the parabola
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Main facts about the parabola
Forms of Parabolas
Axis
Directix
Vertex
Focus
Length of latus
rectum
Equations of latus
rectum
𝑦 2 = 4𝑎𝑥
𝑦=0
𝑥 = −𝑎
(0,0)
(𝑎, 0)
4𝑎
𝑦 2 = −4𝑎𝑥
𝑦=0
𝑥=𝑎
(0,0)
(−𝑎, 0)
4𝑎
𝑥 2 = 4𝑎𝑦
𝑥=0
𝑦 = −𝑎
(0,0)
(0, 𝑎)
4𝑎
𝑥 2 = −4𝑎𝑦
𝑥=0
𝑦=𝑎
(0,0)
(0, −𝑎)
4𝑎
𝑥=𝑎
𝑥 = −𝑎
𝑦=𝑎
𝑦 = −𝑎
Focal distance of a point
Let the equation of the parabola be 𝑦 2 = 4𝑎𝑥 and P(𝑥, 𝑦) be a point on it. Then the distance of P from
the focus (𝑎, 0) is called the focal distance of the point, i.e.,
FP == √(𝑥 − 𝑎)2 + 𝑦 2 = √(𝑥 − 𝑎)2 + 4𝑎𝑥 = √(𝑥 + 𝑎)2 = |𝑥 + 𝑎|
Example 5 Find the coordinates of the focus, axis, the equation of the directrix and latus
rectum of the parabola 𝑦 2 = 8𝑥.
Solution The given equation involves 𝑦 2 , so the axis of symmetry is along the 𝑥-axis.
The coefficient of 𝑥 is positive so the parabola opens to the right. Comparing with the given
equation 𝑦 2 = 4𝑎𝑥, we find that 𝑎 = 2.
Thus, the focus of the parabola is (2,0) and the equation of the directrix of the parabola is
𝑥 = −2 Length of the latus rectum is 4𝑎 = 4 × 2 = 8.
Example 6 Find the equation of the parabola with focus (2,0) and directrix 𝑥 = −2.
Solution Since the focus (2,0) lies on the 𝑥-axis, the 𝑥-axis itself is the axis of the parabola.
Hence the equation of the parabola is of the form either 𝑦 2 = 4𝑎𝑥 or 𝑦 2 = −4𝑎𝑥. Since the
directrix is 𝑥 = −2 and the focus is (2,0), the parabola is to be of the form 𝑦 2 = 4𝑎𝑥 with
𝑎 = 2. Hence the required equation is 𝑦 2 = 4(2)𝑥 = 8𝑥
Example 7 Find the equation of the parabola with vertex at (0,0) and focus at (0,2).
Solution Since the vertex is at (0,0) and the focus is at (0,2) which lies on 𝑦-axis, the 𝑦-axis
is the axis of the parabola. Therefore, equation of the parabola is of the form 𝑥 2 = 4𝑎𝑦. thus,
we have 𝑥 2 = 4(2)𝑦, i.e., 𝑥 2 = 8𝑦.
Example 8 Find the equation of the parabola which is symmetric about the 𝑦-axis, and passes
through the point (2, −3).
Solution Since the parabola is symmetric about 𝑦-axis and has its vertex at the origin, the
equation is of the form 𝑥 2 = 4𝑎𝑦 or 𝑥 2 = −4𝑎𝑦, where the sign depends on whether the
parabola opens upwards or downwards. But the parabola passes through (2, −3) which lies in
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the fourth quadrant, it must open downwards. Thus, the equation is of the form 𝑥 2 = −4𝑎𝑦.
Since the parabola passes through (2, −3), we have,
22 = −4𝑎(−3), i.e., 𝑎 =
1
3
1
Therefore, the equation of the parabola is 𝑥 2 = −4 (3) 𝑦, i.e., 3𝑥 2 = −4𝑦
EXERCISE 10.2
In each of the following Exercises 1 to 6, find the coordinates of the focus, axis of the
parabola, the equation of the directrix and the length of the latus rectum.
1
𝑦 2 = 12𝑥.
Sol. Comparing 𝑦 2 = 12𝑥 with 𝑦 2 = 4𝑎𝑥,
we have 4𝑎 = 12 ⇒ 𝑎 = 3
Focus is (𝑎, 0) = (3,0)
Axis is 𝑥-axis.
Equation of the directrix is 𝑥 = −𝑎 i.e., 𝑥 = −3.
Length of the latus rectum = 4𝑎 = 4 × 3 = 12.
2. 𝑥 2 = 6𝑦.
Sol. The given equation is 𝑥 2 = 6𝑦
It is of the form 𝑥 2 = 4𝑎𝑦
6
3
Here 4𝑎 = 6 ∴ 𝑎 = 4 = 2
3
The co-ordinates of the focus are (0, 𝑎) = (0, 2)
3
Equation of directrix is 𝑦 = −𝑎 i.e., 𝑦 = − 2
Length of latus rectum = 4𝑎 = 6
3. 𝑦 2 = −8𝑥.
Sol. Comparing 𝑦 2 = −8𝑥 with 𝑦 2 = −4𝑎𝑥,
we have,
4𝑎 = 8 ⇒ 𝑎 = 2
Focus is (−𝑎, 0) = (−2,0)
Axis is 𝑥-axis.
Equation of directrix is 𝑥 = 𝑎, i.e., 𝑥 = 2.
Length of latus rectum = 4𝑎 = 4 × 2 = 8.
4. 𝑥 2 = −16𝑦.
Sol. The given equation is 𝑥 2 = −16𝑦
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It is of the form
𝑥 2 = −4𝑎𝑦
(Standard Form IV)
Here, 4𝑎 = 16 ∴ 𝑎 = 4
The co-ordinates of the focus are (0, −𝑎) = (0, −4)
Equation of directrix is y = 𝑎 i.e., 𝑦 = 4
Length of latus rectum 4𝑎 = 16
5
𝑦 2 = 10𝑥.
Sol. Comparing 𝑦 2 = 10𝑥 with 𝑦 2 = 4𝑎𝑥, [Standard Form I]
5
we have 4𝑎 = 10 ⇒ 𝑎 = 2
5
Focus is (𝑎, 0) = (2 , 0)
Axis is 𝑥-axis.
𝟓
Equation of directrix is 𝑥 = −𝑎, i.e., 𝑥 = − 2
5
Length of latus rectum = 4𝑎 = 4 × 2 = 10.
6. 𝑥 2 = −9𝑦.
Sol. Comparing 𝑥 2 = −9𝑦 with 𝑥 2 = −4𝑎𝑦, [Standard Form IV]
9
we have 4𝑎 = 9 ⇒ 𝑎 = 4
9
Focus is (0, −𝑎) = (0, − 4)
Axis is 𝑦-axis.
9
Equation of directrix is 𝑦 = 𝑎, i.e., 𝑦 = 4.
9
Length of latus rectum = 4𝑎 = 4 × 4 = 9.
In each of the following Exercises 7 to 12, find the equation of the parabola that satisfies
the given conditions:
7. Focus (6,0); directrix 𝑥 = −6.
Sol. Since the focus (6,0) lies on the 𝑥-axis, (∵ 𝑦 = 0), the 𝑥-axis itself is the axis of the
parabola. Therefore, the equation of the parabola is either
𝑦 2 = 4𝑎𝑥 or 𝑦 2 = −4𝑎𝑥 (𝑎 > 0)
(Standard Form I or II)
Since the focus (6,0) lies to the right of the vertex (0,0); the parabola is right hand parabola
𝑦 2 = 4𝑎𝑥.
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Since focus (𝑎, 0) is (6,0). Comparing 𝑎 = 6
Hence, the required equation of parabola is 𝑦 2 = 4𝑎𝑥 or 𝑦 2 = 4(6)𝑥 or 𝑦 2 = 24𝑥.
8. Focus (0, −3); directrix 𝑦 = 3.
Sol. Since the focus (0, −3) lies on the 𝑦-axis, (∵ 𝑥 = 0), the 𝑦 axis itself is the axis of the
parabola. Therefore,
the equation of the parabola is either 𝑥 2 = 4𝑎𝑦 or 𝑥 2 = −4𝑎𝑦 (𝑎 > 0)
Since the focus (0, −3) lies below the origin and directrix is 𝑦 = 3, the parabola is
downward parabola 𝑥 2 = −4𝑎𝑦 with 𝑎 = 3. Hence the required equation is
𝑥 2 = −4(3)𝑦 or 𝑥 2 = −12𝑦.
9
Vertex (0,0); focus (3,0).
Sol. Since the focus (3,0) lies on the 𝑥-axis, the 𝑥-axis itself is the axis of the parabola.
Therefore, the equation of the parabola is
either 𝑦 2 = 4𝑎𝑥 or 𝑦 2 = −4𝑎𝑥.
Since the focus (3,0) lies on the right of the origin, the parabola is right hand parabola 𝑦 2 =
4𝑎𝑥 with 𝑎 = 3.
Hence the required equation is 𝑦 2 = 4(3)𝑥 or 𝑦 2 = 12𝑥.
10. Vertex (0,0); focus (−2,0).
Sol. Since the focus (−2,0) lies on the 𝑥-axis, the 𝑥-axis itself is the axis of the parabola.
Therefore, the equation of the parabola is either 𝑦 2 = 4𝑎𝑥 or 𝑦 2 = −4𝑎𝑥.
Since the focus (−2,0) lies on the left of the origin,
the parabola is left hand parabola 𝑦 2 = −4𝑎𝑥 with 𝑎 = 2.
Hence the required equation is 𝑦 2 = −4(2)𝑥 or 𝑦 2 = −8𝑥.
11 Vertex (0,0); passing through (2,3) and axis is along 𝒙-axis.
Sol. Since the axis of the parabola is along 𝑥-axis, its equation is of the form 𝑦 2 = 4𝑎𝑥 or
𝑦 2 = −4𝑎𝑥, where the sign depends on whether the parabola opens rightwards or
leftwards. Since the parabola passes through (2,3) which lies in the first quadrant, it must
open rightwards. Thus the equation is of the form 𝑦 2 = 4𝑎𝑥
Since it passes through (2,3), we have
9
32 = 4𝑎(2) ⇒ 𝑎 = 8
9
9
9
Putting 𝑎 = 8 in (𝑖), the required equation is 𝑦 2 = 4 (8) 𝑥 or 𝑦 2 = 2 𝑥
or 2𝑦 2 = 9𝑥.
12 Vertex (0,0), passing through (5,2) and symmetric with respect to 𝒚-axis.
Sol. Since the parabola is symmetric about 𝑦-axis and has its vertex at the origin, the
equation is of the form 𝑥 2 = 4𝑎𝑦 or 𝑥 2 = −4𝑎𝑦, where the sign depends on whether the
parabola opens upwards or downwards. But the parabola passes through (5,2) which lies in
the first quadrant, it must open upwards.
Thus the equation is of the form 𝑥 2 = 4𝑎𝑦 … (𝑖)
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25
Since it passes through (5,2), we have 52 = 4𝑎(2) ⇒ 𝑎 = 8
25
25
25
Putting 𝑎 = 8 in (𝑖), the required equation is 𝑥 2 = 4 ( 8 ) 𝑦 or 𝑥 2 = 2 𝑦
or 2𝑥 2 = 25𝑦.
Ellipse: An ellipse is the set of points in a plane, the sum of whose distances from two fixed
points is constant. Alternatively, an ellipse is the set of all points in the plane whose distances
from a fixed point in the plane bears a constant ratio, less than, to their distance from a fixed
line in the plane. The fixed point is called focus, the fixed line a directrix and the constant
ratio (𝑒) the centricity of the ellipse.
We have two standard forms of the ellipse, i.e.,
𝑥2
𝑦2
(i) 𝑎2 + 𝑏2 = 1 and
𝑥2
𝑦2
(ii) 𝑏2 + 𝑎2 = 1,
In both cases 𝑎 > 𝑏 and 𝑏 2 = 𝑎2 (1 − 𝑒 2 ), 𝑒 < 1.
In (𝑖) major axis is along 𝑥-axis and minor along 𝑦-axis and in (ii) major axis is along 𝑦 axis
and minor along 𝑥-axis as shown in Fig. 11.8 (a) and (b) respectively.
Focal Distance
𝑥2
𝑦2
The focal distance of a point (𝑥, 𝑦) on the ellipse 𝑎2 + 𝑏2 = 1 is
(𝑖) 𝑎 − 𝑒|𝑥| from the nearer focus (ii) 𝑎 + 𝑒|𝑥| from the farther focus
Sum of the focal distances of any point on an ellipse is constant and equal to the length of the
major axis = 2a
Relationship between semi-major axis, semiminor axis and the distance of the focus from
the centre of the ellipse
Take a point 𝑃 at one end of the major axis. 𝐑 Sum
of the distances of the point 𝑃 to the foci is F1 P +
F2 P = F1 O + OP + F2 P
(Since, F1 P = F1 O + OP )
= 𝑐 + 𝑎 + 𝑎 − 𝑐 = 2𝑎
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Take a point 𝑄 at one end of the minor axis.
Sum of the distances from the point 𝑄 to the foci is
F1 Q + F2 Q = √𝑏 2 + 𝑐 2 + √𝑏 2 + 𝑐 2 = 2√𝑏 2 + 𝑐 2
Since both 𝑃 and 𝑄 lies on the ellipse.
By the definition of ellipse, we have
= 2𝑎, i.e., 𝑎 = √𝑏 2 + 𝑐 2
2√𝑏 2 + 𝑐 2
𝑎2
= 𝑏 2 + 𝑐 2 , i.e., 𝑐 = √𝑎2 − 𝑏 2
𝑎2 = 𝑏 2 + 𝑐 2 , i.e., 𝑐 = √𝑎2 − 𝑏 2 .
Or
Eccentricity
Definition 5 The eccentricity of an ellipse is the ratio of the distances from the centre of the
ellipse to one of the foci and to one of the vertices of the ellipse (eccentricity is denoted by 𝑒 )
𝑐
i.e., 𝑒 = 𝑎. Then since the focus is at a distance of 𝑐 from the centre, in terms of the
eccentricity the focus is at a distance of 𝑎𝑒 from the centre.
Latus rectum
Definition Latus rectum of an ellipse is a line segment
perpendicular to the major axis through any of the foci
and whose end points lie on the ellipse (Fig 10.26).
𝑥2
To find the length of the latus rectum of the ellipse 𝑎2 +
𝑦2
𝑏2
=1
Let the length of AF2 be 𝑙.
Fig 10.26
Then the coordinates of A are (𝑐, 𝑙),i.e., (𝑎𝑒, 𝑙)
𝑥2
𝑦2
Since A lies on the ellipse 𝑎2 + 𝑏2 = 1,
we have
(𝑎𝑒)2
𝑎2
𝑙2
+ 𝑏2 = 1
⇒ 𝑙 2 = 𝑏 2 (1 − 𝑒 2 )
But
𝑐2
𝑒 2 = 𝑎2 =
𝑎2 −𝑏2
𝑎2
𝑏4
𝑏2
= 1 − 𝑎2
𝑏2
Therefore, 𝑙 2 = 𝑎2 , i.e., 𝑙 = 𝑎
Since the ellipse is symmetric with respect to 𝑦-axis (of course, it is symmetric w.r.t.
2𝑏 2
both the coordinate axes), AF2 = F2 B and so length of the latus rectum is 𝑎 .
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Main facts about the Ellipse
𝑥2 𝑦2
+
=1
𝑎2 𝑏 2
𝑎>𝑏
𝑦=0
𝑥=0
2𝑏
𝑎
𝑥=±
𝑒
𝑥 = ±𝑎𝑒
2𝑏 2
𝑎
(0,0)
Forms of the ellipse
Equation of major axis
Length of major axis
Equation of Minor axis
Length of Minor axis
Directrices
Equation of latus rectum
Length of latus rectum
Centre
𝑥2 𝑦2
+
=1
𝑏 2 𝑎2
𝑎>𝑏
𝑥=0
2𝑎
𝑦=0
𝑎
𝑦=±
𝑒
𝑦 = ±𝑎𝑒
2𝑏 2
𝑎
(0,0)
Example 9 Find the coordinates of the foci, the vertices, the length of major axis, the minor
axis, the eccentricity and the latus rectum of the ellipse
𝑥2
𝑦2
+ 9 =1
25
𝑥2
𝑦2
Solution Since denominator of 25 is larger than the denominator of 9 , the major
𝑥2
𝑦2
axis is along the 𝑥-axis. Comparing the given equation with 𝑎2 + 𝑏2 = 1,
we get 𝑎 = 5 and 𝑏 = 3. Also 𝑐 = √𝑎2 − 𝑏 2 = √25 − 9 = 4
Therefore, the coordinates of the foci are (−4,0) and (4,0), vertices are (−5,0) and (5,0).
Length of the major axis is 10 units length of the minor axis 2𝑏 is 6 units and the
2𝑏 2
4
18
eccentricity is 5 and latus rectum is 𝑎 = 5 .
Example 10 Find the coordinates of the foci, the vertices, the lengths of major and minor axes
and the eccentricity of the ellipse 9𝑥 2 + 4𝑦 2 = 36.
Solution The given equation of the ellipse can be written in standard form as
𝑥2 𝑦2
+
=1
4
9
𝑦2
𝑥2
Since the denominator of 9 is larger than the denominator of 4 , the major axis is along
the 𝑦-axis. Comparing the given equation with the standard equation
𝑥2 𝑦2
+
= 1, we have 𝑏 = 2 and 𝑎 = 3.
𝑏 2 𝑎2
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𝑐
Also, 𝑐 = √𝑎2 − 𝑏 2 = √9 − 4 = √5
√5
and 𝑒 = 𝑎 = 3
Hence the foci are (0, √5) and (0, −√5), vertices are (0,3) and (0, −3),
length of the major axis is 6 units, the length of the minor axis is 4 units and the
√5
eccentricity of the ellipse is 3 .
Example 11 Find the equation of the ellipse whose vertices are (±13,0) and foci are (±5,0).
Solution Since the vertices are on 𝑥-axis, the equation will be of the form
𝑥2 𝑦2
+
= 1, where 𝑎 is the semi-major axis.
𝑎2 𝑏 2
Given that 𝑎 = 13, 𝑐 = ±5.
Therefore, from the relation 𝑐 2 = 𝑎2 − 𝑏 2 , we get
25 = 169 − 𝑏 2 , i.e., 𝑏 = 12
𝑥2
𝑦2
Hence the equation of the ellipse is 169 + 144 = 1.
Example 12 Find the equation of the ellipse, whose length of the major axis is 20 and foci are (0, ±5).
Solution Since the foci are on 𝑦-axis, the major axis is along the 𝑦-axis. So, equation of the ellipse is
of the form
𝑥2
𝑦2
+ 2 = 1.
2
𝑏
𝑎
Given that 𝑎 = semi-major axis =
20
= 10
2
and the relation 𝑐 2 = 𝑎2 − 𝑏 2 gives 52 = 102 − 𝑏 2 i.e., 𝑏2 = 75
Therefore, the equation of the ellipse is
𝑥2
𝑦2
+ 100 = 1
75
Example 13 Find the equation of the ellipse, with major axis along the 𝑥-axis and passing through the
points (4,3) and (−1,4).
𝑥2
𝑦2
Solution The standard form of the ellipse is 𝑎2 + 𝑏2 = 1. Since the points (4,3) and (−1,4) lie on the
ellipse, we have
16
9
+ 𝑏2 = 1
𝑎2
1
16
+ 𝑏2 = 1
𝑎2
And
… (1)
… (2)
247
247
Solving equations (1) and (2), we find that 𝑎2 = 7 and 𝑏 2 = 15 .
Hence the required equation is
𝑥2
247
( )
7
𝑦2
+ 247 = 1, i.e., 7𝑥 2 + 15𝑦 2 = 247
15
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In each of the Exercises 1 to 9, find the coordinates of the foci, the vertices, the length of
major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
𝑥2
𝑦2
+ 16 = 1.
36
1
𝑥2
𝑦2
Sol. Since the denominator of 36 is larger than the denominator of 16, the major axis is along
the 𝑥-axis.
𝑥2
𝑦2
Comparing the given equation with 𝑎2 + 𝑏2 = 1, we have,
𝑎2 = 36 and 𝑏 2 = 16 so that 𝑎 = 6 and 𝑏 = 4.
Also 𝑐 = √𝑎2 − 𝑏 2 = √36 − 16 = √20 = √4 × 5 = 2√5.
∴ The foci are (±𝑐, 0) = (±2√5, 0)
The vertices are (±𝑎, 0) = (±6,0)
The length of major axis = 2𝑎 = 2 × 6 = 12.
The length of minor axis = 2𝑏 = 2 × 4 = 8.
𝑐
Eccentricity 𝑒 = 𝑎 =
2√5
6
√5
= 3.
2𝑏 2
Length of latus rectum = 𝑎 =
𝑥2
2×16
6
16
= 3.
𝑦2
2. 4 + 25 = 1.
𝑦2
𝑥2
Sol. Since the denominator of 25 is larger than the denominator of 4 , the major axis is
along the 𝑦-axis.
𝑥2
𝑦2
Comparing the given equation with 𝑏2 + 𝑎2 = 1, we have 𝑎2 = 25 and 𝑏 2 = 4 so that 𝑎 =
5 and 𝑏 = 2.
Also 𝑐 = √𝑎2 − 𝑏 2 = √25 − 4 = √21.
∴ The foci are (0, ±𝑐) = (0, ±√21)
The vertices are (0, ±𝑎) = (0, ±5).
The length of major axis = 2𝑎 = 2 × 5 = 10.
The length of minor axis = 2𝑏 = 2 × 2 = 4.
𝑐
Eccentricity 𝑒 = 𝑎 =
√21
.
5
2𝑏 2
Length of latus rectum = 𝑎 =
3
𝑥2
2×4
5
8
= 5.
𝑦2
+ 9 = 1.
16
𝑥2
𝑦2
Sol. Since the denominator of 16 is larger than the denominator of 9 , the major axis is
𝑥2
𝑦2
along the 𝑥-axis. Comparing the given equation with 𝑎2 + 𝑏2 = 1, we have 𝑎2 = 16 and
𝑏 2 = 9 so that 𝑎 = 4 and 𝑏 = 3.
Also 𝑐 = √𝑎2 − 𝑏 2 = √16 − 9 = √7.
∴ The foci are (±𝑐, 0) = (±√7, 0)
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The vertices are (±𝑎, 0) = (±4,0).
The length of major axis = 2𝑎 = 2 × 4 = 8.
The length of minor axis = 2𝑏 = 2 × 3 = 6.
𝑐
√7
Eccentricity 𝑒 = 𝑎 = 4 .
2𝑏 2
Length of latus rectum = 𝑎 =
4
𝑥2
2×9
4
9
= 2.
𝑦2
+ 100 = 1
25
𝑦2
𝑥2
Sol. Since the denominator of 100 is larger than the denominator of 25, the major axis is
𝑥2
𝑦2
along the 𝑦-axis. Comparing the given equation with 𝑏2 + 𝑎2 = 1, we have
𝑎2 = 100 and 𝑏 2 = 25 so that 𝑎 = 10 and 𝑏 = 5.
Also 𝑐 = √𝑎2 − 𝑏 2 = √100 − 25 = √75 = √25 × 3 = 5√3.
∴ The foci are (0, ±𝑐) = (0, ±5√3)
The vertices are (0, ±𝑎) = (0, ±10).
The length of major axis = 2𝑎 = 2 × 10 = 20.
The length of minor axis = 2𝑏 = 2 × 5 = 10.
𝑐
5√3
√3
Eccentricity 𝑒 = 𝑎 = 10 = 2 .
2𝑏 2
Length of latus rectum = 𝑎 =
𝑥2
10
= 5.
𝑦2
+ 36 = 1.
49
5.
2×25
𝑥2
𝑦2
Sol. Since the denominator of 49 is larger than the denominator of 36, the major axis is along
the 𝑥-axis.
𝑥2
𝑦2
Comparing the given equation with 𝑎2 + 𝑏2 = 1, we have 𝑎2 = 49 and 𝑏 2 = 36 so that 𝑎 =
7 and 𝑏 = 6.
Also 𝑐 = √𝑎2 − 𝑏 2 = √49 − 36 = √13.
∴ The foci are (±𝑐, 0) = (±√13, 0)
The vertices are (±𝑎, 0) = (±7,0).
The length of major axis = 2𝑎 = 2 × 7 = 14.
The length of minor axis = 2𝑏 = 2 × 6 = 12.
𝑐
Eccentricity 𝑒 = 𝑎 =
√13
.
7
2𝑏 2
Length of latus rectum = 𝑎 =
𝑥2
2×36
7
72
= 7.
𝑦2
6. 100 + 400 = 1.
𝑦2
𝑥2
Sol. Since the denominator of 400 is larger than the denominator of 100, the major axis is
𝑥2
𝑦2
along the 𝑦-axis. Comparing the given equation with 𝑏2 + 𝑎2 = 1, we have 𝑎2 = 400 and
𝑏 2 = 100 so that 𝑎 = 20 and 𝑏 = 10.
Also 𝑐 = √𝑎2 − 𝑏 2 = √400 − 100 = √300 = √100 × 3 = 10√3.
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∴ The foci are (0, ±𝑐) = (0, ±10√3)
The vertices are (0, ±𝑎) = (0, ±20).
The length of major axis = 2𝑎 = 2 × 20 = 40.
The length of minor axis = 2𝑏 = 2 × 10 = 20.
𝑐
Eccentricity 𝑒 = 𝑎 =
10√3
20
√3
= 2.
2𝑏 2
2×100
Length of latus rectum = 𝑎 =
20
= 10.
7. 36𝑥 2 + 4𝑦 2 = 144.
Sol. Dividing by 144 (to make the R.H.S. unity), the given equation becomes
𝑥2 𝑦2
+
=1
4 36
𝑦2
𝑥2
Since the denominator of 36 is larger than the denominator of 4 , the major axis is along the
𝑦-axis.
𝑥2
𝑦2
Comparing with 𝑏2 + 𝑎2 = 1, we have
𝑎2 = 36 and 𝑏 2 = 4 so that 𝑎 = 6 and 𝑏 = 2.
Also 𝑐 = √𝑎2 − 𝑏 2 = √36 − 4 = √32 = √16 × 2 = 4√2.
The foci are (0, ±𝑐) = (0, ±4√2)
The vertices are (0, ±𝑎) = (0, ±6).
The length of major axis = 2𝑎 = 2 × 6 = 12.
The length of minor axis = 2𝑏 = 2 × 2 = 4.
𝑐
Eccentricity 𝑒 = 𝑎 =
4√2
6
=
2√2
2𝑏 2
3
.
Length of latus rectum = 𝑎 =
2×4
6
4
= 3.
8. 16𝑥 2 + 𝑦 2 = 16.
Sol. Dividing by 16 , (to make the R.H.S. unity,)
𝑥2
𝑦2
+ 16 = 1
1
the given equation becomes
𝑦2
𝑥2
Since the denominator of 16 is larger than the denominator of 1 , the major axis is along the
𝑥2
𝑦2
𝑦-axis. Comparing with 𝑏2 + 𝑎2 = 1, we have
𝑎2 = 16 and 𝑏 2 = 1 so that 𝑎 = 4 and 𝑏 = 1.
Also 𝑐 = √𝑎2 − 𝑏 2 = √16 − 1 = √15.
∴ The foci are (0, ±𝑐) = (0, ±√15)
The vertices are (0, ±𝑎) = (0, ±4).
The length of major axis = 2𝑎 = 2 × 4 = 8.
The length of minor axis = 2𝑏 = 2 × 1 = 2.
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MANGAF (KUWAIT)
𝑐
Eccentricity 𝑒 = 𝑎 =
√15
.
4
2𝑏 2
Length of latus rectum = 𝑎 =
2×1
4
1
= 2.
9. 4𝑥 2 + 9𝑦 2 = 36.
Sol. Dividing by 36 , (to make the R.H.S. unity)
the given equation becomes
𝑥2
𝑦2
+ 4 =1
9
𝑥2
𝑦2
Since the denominator of 9 is larger than the denominator of 4 , the major axis is along the
𝑥2
𝑦2
𝑥-axis. Comparing with 𝑎2 + 𝑏2 = 1, we have 𝑎2 = 9 and 𝑏 2 = 4 so that 𝑎 = 3 and 𝑏 = 2.
Also 𝑐 = √𝑎2 − 𝑏 2 = √9 − 4 = √5.
∴ The foci are (±𝑐, 0) = (±√5, 0)
The vertices are (±𝑎, 0) = (±3,0).
The length of major axis = 2𝑎 = 2 × 3 = 6.
The length of minor axis = 2𝑏 = 2 × 2 = 4.
𝑐
√5
Eccentricity e == 𝑎 = 3
2𝑏 2 2 × 4 8
Length of latus rectum =
=
= .
𝑎
3
3
In each of the following Exercises 10 to 20, find the equation for the ellipse that satisfies the
given conditions:
10. Vertices (±5,0), foci (±4,0).
Sol. Since the vertices and foci are on 𝑥-axis (∵ 𝑦 = 0), therefore major axis is along 𝑥-axis
and hence equation of ellipse is of the form
𝑥2
𝑎2
𝑦2
+ 𝑏2 = 1
Vertices are (±𝑎, 0) = (±5,0)
(given) ∴ 𝑎 = 5 ∴ 𝑎2 = 25
Foci are (±𝑐, 0) = (±4,0) (given) ∴ 𝑐 = 4
The relation c 2 = 𝑎2 − 𝑏 2 gives 42 = 52 − 𝑏 2
⇒ 𝑏 2 = 25 − 16 = 9
Putting these values of 𝑎2 and 𝑏 2 in (i),
Equation of ellipse is
𝑥2 𝑦2
9𝑥 2 + 25𝑦 2
+
= 1 or
= 1 or 9𝑥 2 + 25𝑦 2 = 225.
25 9
25 × 9
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I INDIA INTERNATIONAL SCHOOL
Class XI
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11 Vertices (0, ±13), foci (0, ±5).
Sol. Since the vertices and foci are on 𝑦-axis (∵ 𝑥 = 0), therefore, major axis is along 𝑦axis and hence equation of ellipse is of the form
𝑥2
𝑦2
+ 𝑎2 = 1
𝑏2
… (𝑖)
Vertices are (0, ±𝑎) = (0, ±13) ⇒ 𝑎 = 13
Foci are (0, ±𝑐) = (0, ±5) ⇒ 𝑐 = 5
The relation 𝑐 2 = 𝑎2 − 𝑏 2 gives 52 = (13)2 − 𝑏 2
⇒ 𝑏 2 = 169 − 25 = 144.
∴ From (i), the equation of ellipse is
𝑥2
𝑦2
+ 169 = 1
144
12 Vertices (±6,0), foci (±4,0).
Sol. Since the vertices and foci are on 𝑥-axis (∵ 𝑦 = 0), major axis is along 𝑥-axis and
equation of ellipse is of the form
𝑥2
𝑎2
𝑦2
… (𝑖)
+ 2=1
𝑏
Vertices are (±𝑎, 0) = (±6,0)
Foci are (±𝑐, 0) = (±4,0)
⇒ 𝑎=6
⇒𝑐=4
The relation 𝑐 2 = 𝑎2 − 𝑏 2 gives 42 = 62 − 𝑏 2
⇒ 𝑏 2 = 36 − 16 = 20.
𝑥2
∴ From (i), the equation of ellipse is
𝑦2
+ 20 = 1
36
13 Ends of major axis (±3,0), ends of minor axis (0, ±2).
Sol. Since the major axis is along 𝑥-axis ( ∵ 𝑦 = 0), the equation of ellipse is of the form
𝑥2
𝑦2
+ 𝑏2 = 1
𝑎2
… (𝑖)
Ends of major axis are (±𝑎, 0) = (±3,0) ⇒ 𝑎 = 3
Ends of minor axis are (0, ±𝑏) = (0, ±2) ⇒ 𝑏 = 2
∴ From (i), the equation of ellipse is
𝑥2
𝑦2
+ 4 = 1.
9
14 Ends of major axis (0, ±√5), ends of minor axis (±1,0).
Sol. Since the major axis is along 𝑦-axis ( ∵ 𝑥 = 0), the equation of ellipse is of the form
𝑥2
𝑦2
+ 𝑎2 = 1 … (𝑖)
𝑏2
Ends of major axis are (0, ±𝑎) = (0, ±√5) ⇒ 𝑎 = √5
Ends of minor axis are (±𝑏, 0) = (±1,0) ⇒ 𝑏 = 1
∴ From (i), the equation of ellipse is
𝑥2
𝑦2
+ 5 =1
1
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I INDIA INTERNATIONAL SCHOOL
Class XI
MANGAF (KUWAIT)
15 Length of major axis 26, foci (±5,0).
Sol. Since the foci are on 𝑥-axis ( ∵ 𝑦 = 0), the major axis is along 𝑥-axis and the equation of
ellipse is of the form
𝑥2
𝑦2
+ 𝑏2 = 1 … (𝑖)
𝑎2
Length of major axis is 2𝑎 = 26 ⇒ 𝑎 = 13
Foci are (±𝑐, 0) = (±5,0) ⇒ 𝑐 = 5
The relation 𝑐 2 = 𝑎2 − 𝑏 2 gives 52 = (13)2 − 𝑏 2
⇒ 𝑏 2 = 169 − 25 = 144
∴ From (i), the equation of ellipse is
𝑥2
169
+
𝑦2
144
=1
16 Length of minor axis 16 , foci (0, ±6).
Sol. Since the foci are on 𝑦-axis (∵ 𝑥 = 0), the major axis is along 𝑦-axis and the equation
𝑥2
𝑦2
+ 𝑎2 = 1
𝑏2
of ellipse is of the form
… (𝑖)
Length of minor axis is 2𝑏 = 16 ⇒ 𝑏 = 8
Foci are (0, ±𝑐) = (0, ±6) ⇒ 𝑐 = 6
The relation 𝑐 2 = 𝑎2 − 𝑏 2 gives 62 = 𝑎2 − 82
⇒ 𝑎2 = 36 + 64 = 100
∴ From (i), the equation of ellipse is
𝑥2
𝑦2
+ 100 = 1
64
17. Foci (±3,0), 𝑎 = 4.
Sol. Since the foci are on 𝑥-axis ( ∵ 𝑦 = 0), the major axis is along 𝑥-axis and the equation
𝑥2
𝑦2
of ellipse is of the form𝑎2 + 𝑏2 = 1 … (𝑖)
Foci are (±𝑐, 0) = (±3,0) ⇒ 𝑐 = 3
Also 𝑎 = 4 (given)
The relation 𝑐 2 = 𝑎2 − 𝑏 2 gives 32 = 42 − 𝑏 2
⇒ 𝑏 2 = 16 − 9 = 7
∴ From (i), the equation of ellipse is
𝑥2
𝑦2
+ 7 =1
16
18 𝑏 = 3, 𝑐 = 4, centre at the origin, foci on the 𝑥-axis.
Sol. Because centre is at the origin and foci are on 𝑥-axis, therefore,
equation of ellipse is
𝑥2
𝑦2
+ 𝑏2 = 1 (𝑖)
𝑎2
b = 3, c= 4 (given)
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I INDIA INTERNATIONAL SCHOOL
Class XI
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We know that for ellipse 𝑎2 = 𝑏 2 + 𝑐 2 = 32 + 42 = 25
Putting values of 𝑎2 and 𝑏 2 in ( 𝑖),
required equation of ellipse is
𝑥2
𝑦2
+ 9 = 1.
25
19 Centre at (0,0), major axis on the 𝑦-axis and passes through the points (3,2) and (1,6).
Sol. Since the major axis is on the 𝑦-axis,
𝑥2
the equation of ellipse is of the form
𝑏2
𝑦2
+ 𝑎2 = 1
… (𝑖)
The points (3,2) and (1,6) lie on (𝑖)
9
4
∴ 𝑏2 + 𝑎2 = 1 (ii)
1
𝑏2
36
+ 𝑎2 = 1
(iii)
Multiplying (iii) by 9, we have
9
𝑏2
324
… (iv)
+ 𝑎2 = 9
Subtracting (ii) from (iv) (to eliminate 𝑏 2 ), we have
320
= 8 ⇒ 8𝑎2 = 320 ⇒ 𝑎2 = 40.
𝑎2
1
36
Putting 𝑎2 = 40 in (iii), we have 𝑏2 + 40 = 1
⇒
1
9
1
= 1−
=
⇒ 𝑏 2 = 10
2
𝑏
10 10
𝑥2
𝑦2
∴ From (i), the equation of ellipse is10 + 40 = 1 or 4𝑥 2 + 𝑦 2 = 40
20 Major axis on the 𝑥-axis and passes through the points (4,3) and (6,2).
Sol. Since the major axis is on the 𝑥-axis, the equation of ellipse is of the form
𝑥2
𝑦2
… (𝑖)
+ 𝑏2 = 1
𝑎2
Since the points (4,3) and (6,2) lie on (𝑖)
16
9
∴
+ 𝑏2 = 1
… (ii)
𝑎2
and
36
𝑎2
4
+ 𝑏2 = 1
… (iii)
Multiplying (ii) by 4 , (iii) by 9 and subtracting (to eliminate 𝑏 2 , we get
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I INDIA INTERNATIONAL SCHOOL
Class XI
MANGAF (KUWAIT)
64
𝑎2
324
− 𝑎2 = 4 − 9 ⇒
⇒
−260
𝑎2
64−324
𝑎2
= −5,
260
= −5 ⇒ 5𝑎2 = 260, ⇒ 𝑎2 = 5 = 52
16
9
Putting 𝑎2 = 52 in (ii), we have52 + 𝑏2 = 1
⇒
9
4
9
=
1
−
=
⇒ 9𝑏 2 = 9 × 13 ⇒ 𝑏 2 = 13
𝑏2
13 13
∴ From (i), the equation of ellipse is
𝑥2 𝑦2
𝑥 2 + 4𝑦 2
+
= 1 or
= 1 or 𝑥 2 + 4𝑦 2 = 52.
52 13
52
Hyperbola
Definition. A
hyperbola is the set of
all points in a plane, the
difference of whose
distances from two
fixed points in the
plane is a constant.
The term “difference”
that is used in the
definition means the
distance to the
farther point minus the
distance to the closer
point. The two fixed points are called the foci of the hyperbola. The mid-point of the line
segment joining the foci is called the centre of the hyperbola. The line through the foci is
called the transverse axis and the line through the centre and perpendicular to the transverse
axis is called the conjugate axis. The points at which the hyperbola
intersects the transverse axis are called the vertices of the hyperbola (Fig).
We denote the distance between the two foci by 2c, the distance between two vertices (the
length of the transverse axis) by 2a and we define the quantity b as
𝑏 = √𝑐 2 − 𝑎 2
Also 2𝑏 is the length of the conjugate axis as shown in(fig).
To find the constant 𝑷𝟏 𝑭𝟐 − 𝑷𝟏 𝑭𝟏 :
By taking the point P at A and B in the Fig 10.28, we have BF1 − BF2 = AF2 − AF1 (by the
definition of the hyperbola) BA + AF1 − BF2 = AB + BF2 − AF1 i.e., AF1 = BF2
So that, BF1 − BF2 = BA + AF1 − BF2 = BA = 2𝑎
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MANGAF (KUWAIT)
c
Eccentricity e = a where 𝑐 ≥ 𝑎 and In terms of the eccentricity, the foci are at a distance of
ae from the centre
We have two standard forms of the hyperbola, i.e.,
𝑥2
𝑦2
(𝑖) 𝑎2 − 𝑏2 = 1 and
𝑦2
𝑥2
(ii) 𝑎2 − 𝑏2 = 1
Here 𝑏 2 = 𝑎2 (𝑒 2 − 1), 𝑒 > 1.
In (𝑖) transverse axis is along 𝑥-axis and conjugate axis along 𝑦-axis where as in (ii)
transverse axis is along 𝑦-axis and conjugate axis along 𝑥-axis.
Main facts about the Hyperbola
𝑥2 𝑦2
−
=1
𝑎2 𝑏 2
𝑦2 𝑥2
−
=1
𝑎2 𝑏 2
Equation of transverse axis
Equation of conjugate axis
Length of transverse axis
Foci
𝑦=0
𝑥=0
2𝑎
(±𝑎𝑒, 0)
𝑥=0
𝑦=0
2𝑎
(0, ±𝑎𝑒)
Equation of latus rectum
𝑥 = ±𝑎𝑒
Length of latus rectum
2𝑏 2
𝑎
(0,0)
2𝑏 2
𝑎
(0,0)
Forms of the hyperbola
Centre
1
𝑐
Example 14 Find the coordinates of the foci and the vertices, the eccentricity, the length of the latus
rectum of the hyperbolas:
𝑥2
𝑦2
− = 1, (ii) 𝑦 2 − 16𝑥 2 = 16
9
16
𝑥2
𝑦2
Solution (𝑖) Comparing the equation − = 1 with the standard equation
9
16
(𝑖)
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𝑥2 𝑦2
−
=1
𝑎2 𝑏 2
Here, 𝑎 = 3, 𝑏 = 4 and 𝑐 = √𝑎2 + 𝑏 2 = √9 + 16 = 5
Therefore, the coordinates of the foci are (±5,0) and that of vertices are (±3,0).
𝑐
5
2𝑏2
32
𝑦2
𝑥2
Also, The eccentricity 𝑒 = 𝑎 = 3. The latus rectum = 𝑎 = 3
(ii) Dividing the equation by 16 on both sides, we have 16 − 1 = 1
𝑦2
𝑥2
Comparing the equation with the standard equation 𝑎2 − 𝑏2 = 1,
we find that 𝑎 = 4, 𝑏 = 1 and 𝑐 = √𝑎2 + 𝑏 2 = √16 + 1 = √17.
Therefore, the coordinates of the foci are (0, ±√17) and that of the vertices are (0, ±4). Also,
𝑐
The eccentricity 𝑒 = 𝑎 =
2𝑏2
1
√17
.
The
latus
rectum
=
= 2.
4
𝑎
Example 15 Find the equation of the hyperbola with foci (0, ±3) and vertices (0, ±
𝑦2
√11
)
2
𝑥2
Solution Since the foci is on y-axis, the equation of the hyperbola is of the form 𝑎2 − 𝑏2 = 1
√11
Since vertices are (0, ± 2 ) , 𝑎 =
√11
2
Also, since foci are (0, ±3); 𝑐 = 3 and 𝑏 2 = 𝑐 2 − 𝑎2 =
25
.
4
Therefore, the equation of the hyperbola is
𝑦2
𝑥2
−
= 1, i.e., 100𝑦 2 − 44𝑥 2 = 275.
11
25
( ) ( )
4
4
Example 16 Find the equation of the hyperbola where foci are (0, ±12) and the length of the latus
rectum is 36 .
Solution Since foci are (0, ±12), it follows that 𝑐 = 12.
2𝑏 2
Length of the latus rectum = 𝑎 = 36 or 𝑏 2 = 18𝑎
Therefore,
𝑐 2 = 𝑎2 + 𝑏 2 ; gives144 = 𝑎2 + 18𝑎
⇒ 𝑎2 + 18𝑎 − 144 = 0 ⇒ 𝑎 = −24 𝑜𝑟 6
Since 𝑎 cannot be negative, we take 𝑎 = 6 and so 𝑏 2 = 108.
𝑦2
𝑥2
Therefore, the equation of the required hyperbola is 36 − 108 = 1,
i.e., 3𝑦 2 − 𝑥 2 = 108
EXERCISE 10.4
In each of the Exercises 1 to 6 , find the coordinates of the foci and the vertices, the
eccentricity and the length of the latus rectum of the hyperbolas.
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𝑥2
𝑦2
1 16 − 9 = 1.
𝑥2
𝑦2
𝑥2
𝑦2
Sol. Comparing 16 − 9 = 1 with the standard equation 𝑎2 − 𝑏2 = 1, we have
𝑎2 = 16 and 𝑏 2 = 9 so that 𝑎 = 4 and 𝑏 = 3
Also 𝑐 = √𝑎2 + 𝑏 2 = √16 + 9 = 5.
∴ The foci are (±𝑐, 0) = (±5,0)
The vertices are (±𝑎, 0) = (±4,0)
𝑐
5
Eccentricity 𝑒 = 𝑎 = 4
2𝑏 2
Length of latus rectum = 𝑎 =
2.
𝑦2
2×9
4
9
= 2.
𝑥2
− 27 = 1.
9
𝑦2
𝑥2
𝑦2
𝑥2
Sol. Comparing 9 − 27 = 1 with the standard equation 𝑎2 − 𝑏2 = 1, we have
𝑎2 = 9 and 𝑏 2 = 27 so that 𝑎 = 3 and 𝑏 = √27 = √9 × 3 = 3√3
Also 𝑐 = √𝑎2 + 𝑏 2 = √9 + 27 = √36 = 6.
∴ The foci are (0, ±𝑐) = (0, ±6)
The vertices are (0, ±𝑎) = (0, ±3)
𝑐
6
Eccentricity= 𝑎 = 3 = 2
2𝑏 2 2 × 27
Length of latus rectum =
=
= 18
𝑎
3
3
9𝑦 2 − 4𝑥 2 = 36.
Sol. Dividing both sides of the given equation by 36 (to make the R.H.S. unity),
we get
𝑦2
𝑥2
− 9 =1
4
Comparing it with the standard equation
𝑦2
𝑥2
− 𝑏2 = 1, we have
𝑎2
𝑎2 = 4 and 𝑏 2 = 9 so that 𝑎 = 2 and 𝑏 = 3.
Also, 𝑐 = √𝑎2 + 𝑏 2 = √4 + 9 = √13.
∴ The foci are (0, ±𝑐) = (0, ±√13)
The vertices are (0, ±𝑎) = (0, ±2)
𝑐
Eccentricity 𝑒 = 𝑎 =
√13
2
2𝑏 2
Length of the latus rectum = 𝑎 =
2×9
2
= 9.
4. 16𝑥 2 − 9𝑦 2 = 576.
Sol. Dividing both sides of the given equation by 576 (to make the R.H.S. unity),
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𝑥2
𝑦2
we get 36 − 64 = 1
𝑥2
𝑦2
Comparing it with the standard equation 𝑎2 − 𝑏2 = 1,
we have 𝑎2 = 36 and 𝑏 2 = 64 so that 𝑎 = 6 and 𝑏 = 8.
Also, 𝑐 = √𝑎2 + 𝑏 2 = √36 + 64 = √100 = 10.
∴ The foci are (±𝑐, 0) = (±10,0)
The vertices are (±𝑎, 0) = (±6,0)
𝑐
10
5
Eccentricity 𝑒 = 𝑎 = 6 = 3
2𝑏 2
2×64
Length of latus rectum = 𝑎 =
6
64
= 3.
5. 5𝑦 2 − 9𝑥 2 = 36.
Sol. Dividing both sides of the given equation by 36 (to make the R.H.S. unity), we get
5𝑦 2 𝑥 2
𝑦2
𝑥2
= 1 or
−
=1
36 4
36/5 4
Comparing it with the standard equation
𝑦2 𝑥2
−
= 1, we have
𝑎2 𝑏 2
36
6
𝑎2 = 5 and 𝑏 2 = 4 so that 𝑎 =
√5
and 𝑏 = 2
36
56
4×14
Also 𝑐 = √𝑎2 + 𝑏 2 = √ 5 + 4 = √ 5 = √ 5 =
∴ The foci are (0, ±𝑐) = (0, ±
Eccentricity 𝑒 = 𝑎 =
2√14
√5
6
√5
=
Length of the latus rectum =
√5
2√14
The vertices are (0, ±𝑎) = (0, ±
𝑐
2√14
)
√5
6
)
√5
√14
3
2𝑏 2
𝑎
=
2×4
6
√5
=
4√5
3
.
6. 49𝑦 2 − 16𝑥 2 = 784
Sol: Dividing both sides of the given equation by 784 (to make the R.H.S. unity),
we get
𝑦2
𝑥2
− 49 = 1
16
Comparing it with the standard equation
𝑦2
𝑥2
− 𝑏2 = 1,
𝑎2
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we have 𝑎2 = 16 and 𝑏 2 = 49 so that 𝑎 = 4 and 𝑏 = 7
Also, 𝑐 = √𝑎2 + 𝑏 2 = √16 + 49 = √65
∴ The foci are (0, ±𝑐) = (0, ±√65)
The vertices are (0, ±𝑎) = (0, ±4)
𝑐
Eccentricity 𝑒 = 𝑎 =
√65
4
2𝑏 2
Length of the latus rectum = 𝑎 =
2×49
4
49
= 2.
In each of the Exercises 7 to 15, find the equations of the hyperbola satisfying the given
conditions.
7. Vertices (±2,0), foci (±3,0).
Sol. Since the foci are on 𝑥-axis (∵ 𝑦 = 0),
𝑥2
𝑦2
the equation of the hyperbola is of the form 𝑎2 − 𝑏2 = 1 … (𝑖)
Vertices are (±𝑎, 0) = (±2,0) ⇒ 𝑎 = 2
Foci are (±𝑐, 0) = (±3,0) ⇒ 𝑐 = 3
The relation 𝑐 2 = 𝑎2 + 𝑏 2 gives 32 = 22 + 𝑏 2
⇒ 𝑏2 = 9 − 4 = 5
∴ From (i), the equation of hyperbola is
𝑥2
𝑦2
− 5 =1
4
Vertices are (0, ±5), foci (0, ±8).
Sol. Since the foci are on 𝑦-axis (∵ 𝑥 = 0),
8
the equation of the hyperbola is of the form
𝑦2
𝑎2
𝑥2
− 𝑏2 = 1 … (𝑖)
Vertices are (0, ±𝑎) = (0, ±5) ⇒ 𝑎 = 5
Foci are (0, ±𝑐) = (0, ±8) ⇒ 𝑐 = 8
The relation 𝑐 2 = 𝑎2 + 𝑏 2 gives 82 = 52 + 𝑏 2
⇒ 𝑏 2 = 64 − 25 = 39
∴ From (i), the equation of hyperbola is
9
𝑦2
𝑥2
− 39 = 1
25
Vertices (0, ±3), foci (0, ±5).
Sol. Since the foci are on the 𝑦-axis (∵ 𝑥 = 0),
the equation of hyperbola is of the form
𝑦2
𝑎2
𝑥2
− 𝑏2 = 1 … (𝑖)
Vertices are
Foci are (0, ±𝑎) = (0, ±3) ⇒ 𝑎 = 3
(0, ±𝑐) = (0, ±5) ⇒ 𝑐 = 5
The relation 𝑐 2 = 𝑎2 + 𝑏 2 gives 52 = 32 + 𝑏 2
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⇒ 𝑏 2 = 25 − 9 = 16
∴ From (i), the equation of hyperbola is
𝑦2
𝑥2
− 16 = 1
9
10 Foci (±5,0), the transverse axis is of length 8 .
Sol. Because the foci (±5,0) lie on 𝑥-axis ( ∵ 𝑦 = 0 ), the equation of the hyperbola is of
𝑥2
𝑦2
the form 𝑎2 − 𝑏2 = 1 … (𝑖)
Foci are (±5,0) = (±𝑐, 0)
Comparing 𝑐 = 5
Transverse axis is of length 8(= 2𝑎)
∴ 𝑎 = 4 and hence 𝑎2 = 16
The relation 𝑐 2 = 𝑎2 + 𝑏 2 gives 52 = 42 + 𝑏 2
⇒ 25 = 16 + 𝑏 2 ⇒ 𝑏 2 = 9
Putting values of 𝑎2 and 𝑏 2 in (i),
equation of required hyperbola is
𝑥2
𝑦2
− 9 =1
16
11 Foci (0, ±13), the conjugate axis is of length 24 .
Sol. Because the foci (0, ±13) lie on 𝑦-axis (∵ 𝑥 = 0), the equation of the hyperbola is of
𝑦2
𝑥2
the form 𝑎2 − 𝑏2 = 1.
… (𝑖)
Foci are (0, ±13) = (0, ±𝑐) ∴ 𝑐 = 13
conjugate axis is of length = 24 (given)
i.e., 2𝑏 = 24 ∴ 𝑏 = 12 ∴ 𝑏 2 = 144
We know that for hyperbola, 𝑐 2 = 𝑎2 + 𝑏 2
∴ 169 = 𝑎2 + 144 or 𝑎2 = 169 − 144 = 25
𝑦2
𝑥2
Putting values of 𝑎2 and 𝑏 2 in (i), equation of required hyperbola is 25 − 144 = 1
12 Foci (±3√5, 0), the latus rectum is of length 8 .
Sol. Since the foci (±3√5, 0) are on 𝑥-axis (∵ 𝑦 = 0), the equation of the hyperbola is of
𝑥2
𝑦2
… (𝑖)
the form 𝑎2 − 𝑏2 = 1
Foci are (±𝑐, 0) = (±3√5, 0) ∴ 𝑐 = 3√5
2𝑏 2
Length of latus rectum = 𝑎 = 8 or
𝑏 2 = 4𝑎 (ii)
Putting c = 3√5 and 𝑏 2 = 4𝑎 from (ii) in the relation 𝑐 2 = 𝑎2 + 𝑏 2 ,
we have (3√5)2 = 𝑎2 + 4𝑎 or 45 = 𝑎2 + 4𝑎
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or 𝑎2 + 4𝑎 − 45 = 0
or (𝑎 + 9)(𝑎 − 5) = 0 so that 𝑎 = −9,5
Since 𝑎 (length of semi-transverse axis) cannot be negative, reject 𝑎 = −9.
∴ 𝑎 = 5 and hence 𝑎2 = 25.
and therefore from (ii) 𝑏 2 = 4𝑎 = 4 × 5 = 20
𝑥2
𝑦2
∴ from (𝑖), the equation of the required hyperbola is 25 − 20 = 1
4𝑥 2 − 5𝑦 2
or
= 1 or 4𝑥 2 − 5𝑦 2 = 100.
100
13. Foci (±4,0), the latus rectum is of length 12 .
Sol. Since the foci are on 𝑥-axis ( ∵ 𝑦 = 0),
the equation of the hyperbola is of the form
𝑥2
𝑦2
− 𝑏2 = 1
𝑎2
Foci are (±𝑐, 0) = (±4,0) ⇒ 𝑐 = 4
2𝑏 2
Length of latus rectum = 𝑎 = 12 ⇒ 2𝑏 2 = 12𝑎 ⇒ 𝑏 2 = 6𝑎
(ii)
Putting 𝑐 = 4 and 𝑏 2 = 4𝑎 from (ii),
the relation 𝑐 2 = 𝑎2 + 𝑏 2 gives 16 = 𝑎2 + 6𝑎
or 𝑎2 + 6𝑎 − 16 = 0 or (𝑎 + 8)(𝑎 − 2) = 0
so that 𝑎 = −8,2.
Since 𝑎 (length of semi-transverse axis) cannot be negative, reject 𝑎 = −8 ∴ 𝑎 = 2 and
hence 𝑎2 = 4 and therefore from (ii), 𝑏 2 = 6𝑎 = 6 × 2 = 12.
∴ From (i), the equation of required hyperbola is
𝑥2 𝑦2
−
=1
4 12
4
14 Vertices (±7,0), e = 3.
Sol. Since the vertices are on 𝑥-axis (∵ 𝑦 = 0),
the equation of hyperbola is of the form
𝑥2
𝑎2
𝑦2
… (𝑖)
− 𝑏2 = 1
Vertices are (±𝑎, 0) = (±7,0) ⇒ 𝑎 = 7 ⇒ 𝑎2 = 49
𝑒=
4 𝑐 4
28
⇒ = ⇒ 3𝑐 = 4𝑎 = 4 × 7 = 28 ⇒ 𝑐 =
3 𝑎 3
3
From the relation 𝑐 2 = 𝑎2 + 𝑏 2 , we have
28 2
𝑏 2 = 𝑐 2 − 𝑎2 = ( 3 ) − 72 =
784
9
− 49 =
784−441
9
∴ From (i), the equation of required hyperbola is
=
𝑥2
343
9
𝑦2
𝑥2
9𝑦 2
− 343 = 1 or 49 − 343 = 1
49
0
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15 Foci (0, ±√10), passing through (2,3).
Sol. Since the foci are on 𝑦-axis (∵ 𝑥 = 0),
𝑦2
𝑥2
the equation of the hyperbola is of the form 𝑎2 − 𝑏2 = 1 … (𝑖)
Foci are (0, ±𝑐) = (0, ±√10) ∴ 𝑐 = √10
Since the point (2,3) lies on the hyperbola, we have
9
4
−
=1
𝑎2 𝑏 2
9
4
(∵ 𝑐 2 = 𝑎2 + 𝑏 2 , ⇒ 𝑏 2 = 𝑐 2 − 𝑎2 )
or 𝑎2 − 𝑐 2 −𝑎2 = 1
Putting 𝑐 = √10,
9
4
or 𝑎2 − 10−𝑎2 = 1 ⇒
9(10−𝑎2 )−4𝑎2
𝑎2 (10−𝑎2 )
= 1 cross-multiplying, we have
⇒ 9(10 − 𝑎2 ) − 4𝑎2 = 𝑎2 (10 − 𝑎2 )
⇒ 90 − 9𝑎2 − 4𝑎2 = 10𝑎2 − 𝑎4
⇒ 𝑎4 − 23𝑎2 + 90 = 0
Put 𝑎2 = t. Therefore 𝑡 2 − 23𝑡 + 90 = 0
⇒ 𝑡 2 − 18𝑡 − 5𝑡 + 90 = 0 ⇒ 𝑡(𝑡 − 18) − 5(𝑡 − 18) = 0
⇒ (𝑡 − 18)(𝑡 − 5) = 0
∴ Either 𝑡 − 18 = 0 or 𝑡 − 5 = 0
i.e., 𝑎2 = 18 or 𝑎2 = 5(∵ 𝑡 = 𝑎2 ) ∴ 𝑎2 = 18,5
i.e., 𝑡 = 18 or 𝑡 = 5
i.e., 𝑎2 = 18 or 𝑎2 = 5(∵ 𝑡 = 𝑎2 )
∴ 𝑎2 = 18,5
When 𝑎2 = 18, we get 𝑏 2 = 𝑐 2 − 𝑎2 = 10 − 18 = −8 < 0
∴ Rejecting 𝑎2 = 18, we have 𝑎2 = 5 so that
𝑏 2 = 𝑐 2 − 𝑎2 = 10 − 5 = 5
Putting these values of 𝑎2 and 𝑏 2 in (i),
Equation of the required hyperbola is
𝑦2
𝑥2
− 5 = 1 or 𝑦 2 − 𝑥 2 = 5.
5
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Miscellaneous Examples
Example 17 The focus of a parabolic mirror as shown in Fig. is
at a distance of 5 cm from its vertex. If the mirror is 45 cm deep,
find the distance ABas shown in Fig.
Solution Since the distance from the focus to the vertex is 5 cm.
We have, 𝑎 = 5. If the origin is taken at the vertex and the axis of
the mirror lies along the positive 𝑥-axis, the equation of the
parabolic section is 𝑦 2 = 4(5)𝑥 = 20𝑥
Note that:
𝑥 = 45. Thus 𝑦 2 = 900
Therefore, 𝑦 = ±30
Hence, AB = 2𝑦 = 2 × 30 = 60 cm.
Example 18 A beam is supported at its ends by supports which are 12 metres apart, as shown
in figure. Since the load is concentrated at its centre, there is a deflection of 3 cm at the
centre and the deflected beam is in the shape of a parabola. How far from the centre is the
deflection 1 cm ?
Solution Let the vertex be at the
lowest point and the axis vertical.
Let the coordinate axis be chosen
as shown in Fig
. The equation of the parabola takes the form 𝑥 2 = 4𝑎𝑦.
3
3
Since it passes through (6, 100), we have (6)2 = 4𝑎 (100), i.e., 𝑎 =
1
36×100
12
= 300 m
2
Let AB be the deflection of the beam which is 100 m. Coordinates of B are (𝑥, 100).
2
i.e.𝑥 2 = 4 × 300 × 100 = 24 ∴ 𝑥 = √24 = 2√6 metres
Example 19 A rod AB of length 15 cm rests in between two coordinate axes in such a way
that the end point A lies on 𝑥-axis and end point B lies on 𝑦-axis. A point P(𝑥, 𝑦) is taken on
the rod in such a way that AP = 6 cm. Show that the locus of P is an ellipse.
Solution Let AB be the rod making an angle 𝜃 with OX as shown in Fig and P(𝑥, 𝑦) the point
on it such that AP = 6 cm.
Since AB = 15 cm, we have , PB = 9 cm
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From P draw PQ and PR perpendiculars on 𝑦-axis and 𝑥-axis,
𝑥
respectively. From △ PBQ, cos 𝜃 = 9
𝑦
From △ PRA, sin 𝜃 = 6
Since cos2 𝜃 + sin2 𝜃 = 1
𝑥 2
𝑦 2
( ) +( ) =1
9
6
𝑥2
𝑦2
or 81 + 36 = 1
Thus the locus of P is an ellipse
Question 1: If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus.
Solution: The origin of the coordinate plane is taken at the vertex of the parabolic reflector in
such a way that the axis of the reflector is along the
positive 𝑥-axis.
This can be diagrammatically represented as
The equation of the parabola is of the form 𝑦 2 = 4𝑎𝑥
(as it is opening to the right). Since the parabola passes
through point 𝐴(5,10),
102 = 4𝑎 × 5, ⇒ 100 = 20𝑎
100
⇒ 𝑎 = 20 = 5 Therefore,
the focus of the parabola is (𝑎, 0) = (5,0), which is the
mid-point of the diameter.
Hence, the focus of the reflector is at the mid-point of the diameter.
Question 2: An arch is in the form of a parabola with its axis
vertical. The arch is 10 m high and 5 m wide at the base.
How wide is it 2 m from the vertex of the parabola?
Solution: This can be diagrammatically represented as
shown in figure. The origin of the coordinate plane is taken at
the vertex of the arch in such a way that its vertical axis is
along the positive 𝑦-axis. It can be clearly seen that the
5
parabola passes through point (2 , 10)
5 2
25
5
∴ (2) = 4𝑎 × 10 ⇒ 𝑎 = 4×4×10 = 32
5
Therefore, the arch is in the form of a parabola whose equation is 𝑥 2 = 8 𝑦
5
5
5
When 𝑦 = 2 m, 𝑥 2 = 8 × 2, ⇒ 𝑥 2 = 4 , ⇒ x = √4 m
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5
∴ AB = 2 × √4 m = 2 × 1.118 m = 2.23 (approx
Question 3: The cable of a uniformly loaded suspension bridge hangs in the form of a
parabola. The roadway which is horizontal and 100 m long is supported by vertical wires
attached to the cable, the longest wire being 30 m and the shortest being 6 m. Find the length
of a supporting wire attached to the roadway 18 m
from the middle.
Solution: The vertex is at the lowest point of the
cable. The origin of the coordinate plane is taken as
the vertex of the parabola, while its vertical axis is
taken along the positive 𝑦-axis. This can be
diagrammatically represented as shown in figure.
Here, AB and OC are the longest and the shortest
wires, respectively, attached to the cable.
DF is the supporting wire attached to the roadway,
18 m from the middle.
100
Here, AB = 30 m, OC = 6 m and BC = 2 = 50 m
The equation of the parabola is of the form 𝑥 2 = 4𝑎𝑦 (as it is opening upwards).
The coordinates of point 𝐴 are (50,30 − 6) = (50,24)
Since 𝐴(50,24) is a point on the parabola,
(50)2 = 4𝑎 × 24
50×50
625
⇒ 𝑎 = 4×24 = 24
625
⇒ 𝑎 = 24
625
∴ Equation of the parabola is 𝑥 2 = 4 × 24 × 𝑦 ⇒ 6𝑥 2 = 625𝑦
The 𝑥-coordinate of point D is 18 .
Hence, at 𝑥 = 18, 6(18)2 = 625 , ⇒ 𝑦 =
6×18×18
625
⇒ 𝑦 = 3.11 ∴ DE = 3.11 m
DF = DE + EF = 3.11 m + 6 m = 9.11 m
Thus, the length of the supporting wire attached to the roadway 18 m from the middle is
approximately 9.11 m.
Question 4: An arch is in the form of a semi-ellipse. It is 8𝑚 wide and 2𝑚 high at the centre.
Find the height of the arch at a point 1.5 m from one end.
Solution: Since the height and width of the arc from the centre is 2 m and 8 m respectively,
it is clear that the length of the major axis is 8𝑚, while the length of the semi-minor axis is
2 m. The origin of the coordinate plane is taken as the centre of the ellipse, while
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the major axis is taken along the 𝑥-axis. Hence, the semi-ellipse can be diagrammatically
represented as shown in figure alongside.
The equation of the semi-ellipse will be of the form
𝑥2 𝑦2
+
= 1, 𝑦 ≥ 0
𝑎2 𝑏 2
where 𝑎 is the semi-major axis.
Accordingly, 2𝑎 = 8 ⇒ 𝑎 = 4 and 𝑏 = 2
Therefore, the equation of the semi ellipse
𝑥2
𝑦2
is 𝑎2 + 𝑏2 = 1, 𝑦 ≥ 0
Let A be a point on the major axis such that AB = 1.5 m.
Draw AC ⊥ OB. OA = (4 − 1.5)𝑚 = 2.5𝑚
The 𝑥-coordinate of point C is 2.5 .
On substituting the value of 𝑥 with 2.5 in equation (1), we obtain
(2.5)2
16
𝑦2
+ 4 =1
6.25
𝑦2
⇒ 16 + 4 = 1
6.25
⇒ 𝑦 2 = 4 (1 − 16 )
9.75
⇒ 𝑦 2 = 4 × 16
⇒ 𝑦 2 = 2.4275
𝑦 = 1.56
∴ AC = 1.56 m
Thus, the height of the arch at a point 1.5 m from one end is approximately 1.56 m
Question 5: A rod of length 12 cm moves with its ends
always touching the coordinate axes. Determine the
equation of the locus of a point 𝑃 on the rod, which is 3 cm
from the end in contact with the 𝑥-axis.
Solution: Let AB be the rod making an angle 𝜃 with OX and
P(𝑥, 𝑦) be the point on it such that AP = 3 cm.
Then, PB = AB − AP = (12 − 3)cm = 9 cm
[∵ AB = 12 cm]
From P, draw PQ ⊥ OY and PR ⊥ OX
PQ
𝑥
PR
𝑦
In △ PBQ, cos 𝜃 = PB = 9
In △ PRA, sin 𝜃 = PA = 3
Since, sin2 𝜃 + cos 2 𝜃 = 1
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I INDIA INTERNATIONAL SCHOOL
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MANGAF (KUWAIT)
𝑦 2
𝑥 2
𝑥2 𝑦2
( ) + ( ) = 1, ⇒
+
=1
3
9
81 9
𝑥2
𝑦2
Thus, the equation of the locus of point P on the rod is 81 + 9 = 1
Question 6: Find the area of the triangle formed by the lines joining the vertex of the
parabola 𝑥 2 = 12𝑦 to the ends of its latus rectum.
Solution: The given parabola is 𝑥 2 = 12𝑦
On comparing this equation with 𝑥 2 = 4𝑎𝑦,
we obtain 4𝑎 = 12 ⇒ 𝑎 = 3
∴ The coordinates of foci are S(0, 𝑎) = S(0,3)
Let AB be the latus rectum of the given parabola
The given parabola can be roughly drawn as in the figure
alongside.
At 𝑦 = 3, 𝑥 2 = 12 × 3 ⇒ 𝑥 2 = 36 ⇒ 𝑥 = ±6
∴ The coordinates of 𝐴 are (−6,3), while the coordinates
of 𝐵 are (6,3).
Therefore, the vertices of △ 𝑂𝐴𝐵 are 𝑂(0,0), 𝐴(−6,3), and 𝐵(6,3).
1
Area of △ OAB = 2 |0(3 − 3) + (−6)(3 − 0) + 6(0 − 3)| unit 2
=
1
|(−6)(3) + 6(−3)| unit 2
2
=
1
1
| − 36| unit 2 = × 36 unit 2 = 18 unit 2
2
2
Thus, the required area of the triangle is 18 unit 2
Question 7: A man running a racecourse notes that the sum of the distances from the two
flag posts form him is always 10 m and the distance between the flag posts is 8 m. find the
equation of the posts traced by the man.
Solution: Let A and B be the positions of the two flag posts and P(𝑥, 𝑦) be the position of the
man. Accordingly, PA + PB = 10.
We know that if a point moves in a plane in such a way that the sum of its distances from two
fixed points is constant, then the path is an ellipse and this constant value is equal to the
length of the major axis of the ellipse.
Therefore, the path described by the man is an ellipse where the length of the major axis is
10 m, while points A and B are the foci.
Taking the origin of the coordinate plane as the centre of the ellipse,while taking
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I INDIA INTERNATIONAL SCHOOL
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MANGAF (KUWAIT)
the major axis along the 𝑥-axis, the ellipse can be
diagrammatically represented as shown in figure
alongside
The equation of the ellipse will be of the form
𝑥2
𝑎2
𝑦2
+ 𝑏2 = 1 where 𝑎 is the semi-major axis.
Accordingly, 2𝑎 = 10
⇒𝑎=5
Distance between the foci, 2𝑐 = 8 ⇒ 𝑐 = 4
On using the relation 𝑐 = √𝑎2 − 𝑏 2 or 4 = √25 − 𝑏 2
⇒ 16 = 25 − 𝑏 2
⇒ 𝑏 2 = 25 − 16 = 9, ⇒ 𝑏 = 3
𝑥2
𝑦2
Thus, the equation of the path traced by the man is 25 + 9 = 1
Question 8: An equilateral triangle is inscribed in the parabola 𝑦 2 = 4𝑎𝑥, where one vertex
is at the vertex of the parabola. Find the length of the side of the triangle.
Solution: Let OAB be the equilateral triangle inscribed
in parabola 𝑦 2 = 4𝑎𝑥
Let AB intersect the 𝑥-axis at point C.
Let OC = 𝑘
From the equation of the given parabola, we have
𝑦 2 = 4𝑎𝑘
⇒ 𝑦 = ±2√𝑎𝑘
∴ The respective coordinates of points A and B are
(𝑘, 2√𝑎𝑘), and (𝑘, −2√𝑎𝑘)
AB = CA + CB = 2√𝑎𝑘 + 2√𝑎𝑘 = 4√𝑎𝑘
Since OAB is an equilateral triangle, OA2 = AB2
∴ 𝑘 2 + (2√𝑎𝑘)2 = (4√𝑎𝑘)2
⇒ 𝑘 2 + 4𝑎𝑘 = 16𝑎𝑘
⇒ 𝑘 2 = 12𝑎𝑘 ⇒ 𝑘 = 12𝑎
∴ AB = 4√𝑎𝑘 = 4√𝑎 × 12𝑎 = 4√12𝑎2 = 8√3𝑎
Thus, the side of the equilateral triangle inscribed in parabola 𝑦 2 = 4𝑎𝑥 is 8√3𝑎.
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I INDIA INTERNATIONAL SCHOOL
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MANGAF (KUWAIT)
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