Mathematics
Ellipse
Session - 2
Session Objectives
Session Objectives
1. Equation of the tangent in point (x1, y1) form
2. Equation of normal in point (x1, y1) form
3. Equation of tangent and normal in parametric form
4. Number of tangents drawn from a point to an ellipse
5. Director circle
6. Equation of pair of tangents
7. Equation of chord of contact
8. Equation of normal in slope (m) form
9. Number of normals (co-normal points)
10. Equation of chord whose middle point is given
11. Diameter of ellipse
12. Conjugate diameters
Equation of the Tangent in Point
(x1, y1) Form
2
2
x
y
Equation of tangent to ellipse
+
=1
2
2
a
b
at (x1, y1) is given by
xx1
2
a
Slope of tangent = –
b2 x1
2
a y1
+
yy1
2
b
= 1 T = 0
, point of contact =  x1, y1 
Equation of the Tangent in Point
(x1, y1) Form
Working rule for finding T = 0,
replace x2 by xx1, y2 by yy1, 2x by x + x1
2y by y + y1, 2xy by xy1 + x1y
and keeping constant unchanged.
Equation of Normal in Point (x1, y1)
Form
Equation of normal at
given by
x2
a
2
+
y2
2
 x1, y1  to ellipse
= 1 is
b
x – x1 y – y1
a2 x b2 y
2
2
–
= a – b or
=
2
x1
y1
x1/a
y1/b2
 Slope of normal =
a2 y1
b2 x1
Equation of Tangent and Normal in
Parametric () Form
Equation of tangent at
 a cosθ, bsinθ  to
x2
a
2
+
y2
2
=1
b
using point form is
x a cos 
a2
Slope =

ybsin 
b2
 1 or
x
y
cosθ + sinθ = 1
a
b
–b
cot θ, point of contact  a cosθ, b sinθ 
a
with the ellipse.
Equation of Tangent and Normal in
Parametric () Form
Equation of normal at
 acos , bsin  to
x2
2
a

y2
2
1
b
a2 x
b2 y
2
2


a

b
becomes
acos  b sin 
ax
by
or
–
 a2 – b2 or a x sec  bycosec = a2  b2
cos  sin 
Slope =
a
tanθ, foot of normal  a cosθ, b sinθ 
b
Number of Tangents Drawn From a Point
to an Ellipse
Two tangents can be drawn from a point to
an ellipse it may be real or imaginary.
y
O
x
Number of Tangents Drawn From a Point
to an Ellipse


Observation from m2 h2 - a2 - 2hkm + k 2 - b2 = 0
2 2

2
D  4h k  4 h  a
2
k
2
b
2

 h2 k 2

 4a b 

 1
 a2 b2



2 2
h2
k2
(i) Real and distinct if D  0 

 1, i.e. point lies
2
2
a
b
outside the ellipse.
h2
k2
(ii) Real and coincident if D  0 

 1 , i.e. point
2
2
a
b
lies on the ellipse.
h2
k2
(iii) Imaginary if D  0 

 1, i.e. point lies inside
2
2
a
b
the ellipse.
(iv) m1 + m2 =
2hk
2
h a
2
, m1 m2 =
k2  b2
h2  a2
Director Circle
Director circle is the locus of point
of intersection of perpendicular
tangents to the conic.
Equation of director circle of ellipse
Let (h, k) be the point of intersection of tangents to
x2
a
2

y2
b

2
 1 and slope of tangent be m. Then we have

m2 h2  a2  2hkm  k 2  b2  0
Director Circle
m1 m2 
k 2  b2
2
2
h a
if tangents are
perpendicular, then
k 2  b2
2
2
h a
 1, i.e.
h2  k 2  a2  b2
Hence, locus of (h, k) is
x2 + y2 = a2 + b2.
Equation of Pair of Tangents
As we have seen earlier from a point
(h, k) lying outside the ellipse, we
have two real and distinct tangents
possible. Combined equation of
these tangents is given by
2
 x2 y2
  h2 k 2
  hx ky

 2  2  1  2  2  1   2  2  1
a
 a
 a
b
b
b




(h, k)
y
2
or SS1  T , where S, S1,T have
usual meanings.
Note: We can obtain above equation
by eliminating ‘m’ from
y – k = m(x – h) and


m2 h2  a2  2hkm  k2  b2  0
O
x
Equation of Chord of Contact
Equation of chord of contact of point
(h, k) outside the ellipse
hx
a2
+
ky
b2
x2
a2

y2
b2
 1 is
= 1 or T = 0
(h, k)
y
PQ is chord of contact of point (h, k).
Q
P ()
O
x
Equation of Normal Slope (m) Form
Equation of normal of slope m to
x
2
a2
2
+
y
b2
= 1 is y = mx  m

a2  b2

a2 + b2m2
Slope = m, foot of normal is  acosθ, bsinθ 
Number of Normals (Co-normal Points)
Four normals can be drawn from a point
to an ellipse.
Equation of Chord Whose Middle Point
is Given
Equation of chord of
bisected at (h, k) is
x2
2
a
hx
2
a
+
+
y2
2
b
ky
2
=1
– 1=
b
i.e T = S1
where T, S1 have usual meanings.
h2
2
a
+
ky
2
b
– 1,
Diameter of Ellipse
The locus of mid-point of a system of
parallel chords of an ellipse is called
diameter and chords are called its
double ordinates. The end points of
the diameter lying on the ellipse are
called vertices of diameter.
Equation of diameter of ellipse
Let the system of parallel chords be given by
y = mx + c, where ‘m’ is fixed and ‘c’ is a
variable. Let (h, k) be its middle point. Then
equation of chord with middle point at (h, k)
is given by
T  S1 , i.e.
hx
2
a

ky
2
b
–1
h2
2
a

k2
2
b
–1
Diameter of Ellipse
Then its slope is m.
Hence,
–b2h
2
m
a k
 Locus of (h, k) is or
b2 x
2
 m or y =
a y
b2
2
x
a m
which is the required equation of diameter. Note that
diameter passes through the centre of ellipse.
Hence, equation of diameter bisecting the parallel
chords of slope ‘m’ of ellipse
x2
2
a

y2
2
b
 1 is y 
b2
2
a m
x.
Conjugate diameters
Two diameters of an ellipse are said to
be conjugate diameters. If each
bisects the chords parallel to the other.
Condition of conjugate diameters
Let y  m1x and y  m2 x be two conjugate diameters
x2
of
a
2

y2
b
2
 1 (Recall that diameter of ellipse passes
through the centre of the ellipse.). Then diameter
bisecting the chords parallel to y  m1x is given by
y
b2
2
a m1
x which is given as y= m2x.
Conjugate diameters
Then m2 
b2
or m1m2 =
2
a m1
b2
a2
Thus, y  m1x and y  m2 x are conjugate diameters
of ellipse
x2
a
2

y2
b
2
 1 if m1m2 
b2
a
2
.
Note: Major axis and minor axis are conjugate
diameters, as each bisects the chords parallel
to the other but product of their slopes is not
defined.
Properties of conjugate diameters
(i) The eccentric angles of the ends of a
pair of conjugate diameters of an ellipse

differ by .
2
y
(a cos , b sin )R
P (a cos , b sin )
x
O
y = m1x Q
S
y = m2x
Properties of conjugate diameters
(ii)The sum of the squares of any two
conjugate semi-diameters (half of the
diameter) is constant and is given by
sum of squares of semi-axis, i.e.
OP2  OR2  cons tant  a2  b2
Note: That major axis and minor axis are also
conjugate diameters.
(iii)The product of the focal distances of a point on
an ellipse is equal to the square of the semidiameter which is conjugate to the diameter
passing through this point.
Properties of conjugate diameters
(iv) Tangents at the ends of the pair of
conjugate diameters form a
parallelogram, i.e. ABCD is a
parallelogram.
y
+ 
–
2
R
A
P ()
B
O
Q
+ 
C
D
S
+ 3
—
2
x
Properties of conjugate diameters
(v) The area of the parallelogram
formed by the tangents at the ends
of conjugate diameters is constant
and is given by the product of the
axes, i.e. area (ABCD) = 4ab.
y
+ 
–
2
R
A
P ()
D
x
O
B
S
+ 3
—
2
Q
+ 
C
Concyclic Points
Any circle intersects an ellipse in two or four
real points. They are called concyclic points.
If , , g, d be the eccentric angles of four
concyclic points on an ellipse, then
 +  + g + d = 2n, i.e. even multiple of .
Q()
P()
R(g)
S(d)
Class Exercise - 1
The area of the quadrilateral formed by
the tangents at the end points of latus
x2 y2
rectum to the ellipse

 1 is
9
5
27
(a)
sq. units
(b) 9 sq. units
4
(c) 27 sq. units
27
(d)
sq. units
2
Solution
y
A
5
– 2, –
3
(0, 5)
5
2, –
3
D
B
(– 3, 0)
O
–5
– 2, –
3
( 3, 0)
x
–5
2, –
3
C (0,– 5)
End points of latus rectum are given by

b2  
5
 ae, 
   2,   .
a  
3

Solution contd..
Tangents at these points, using point
form, are given by 
2x 5 y

1
9 35
or  2x  3y  9
9
Equation of AD = 2x + 3y = 9  OA  3, OD 
2
Area of the parallelogram ABCD = 4 × Area AOD
1
 4. .AO.OD
2
9
 2.3.  27 sq. units
2
Hence, answer is (c).
Class Exercise - 2
Find the point on x2  2y2  6, which is
nearest to the line x + y = 7.
Solution
y
7 Q
3
P
O
–6
P
6
7
x
–3
The point which is nearest to the line is the point
in the first quadrant, where tangent is parallel to
PQ or if PQ is moved parallel to itself towards
ellipse, the point where PQ first meets, i.e.
touches the ellipse is the required point.
Solution contd..
Let  x1, y1  be the required point,
tangent at is  x1, y1 
xx1  2yy1  6  its slope  1

 x1
 1  x1  2y1
2y1
Also
 x1, y1 lies on
x2  2y2  6
 x12  2y12  6  4y12  2y12  6  y1  1  x1  2
 (2, 1) is the required point.
Note: (–2, –1) is the farthest from the line x + y = 7
Solution contd..
Alternative:
Slope of tangent is –1.
Point of contact is


a2m
b2
,



2 2
2
2 2
2 
a
m

b
a
m

b



a2m
b2
or  –
,–

a2m2  b2
a2m2  b2

  2,  1
 Required point is (2, 1).

 6 1

     ,  3 



6

3
6

3



Class Exercise - 3
If the normal at the end of a latus rectum
of an ellipse of eccentricity ‘e’ passes
through one end of the minor axis, then
e2  e4 
(a) 1
(b) 2
(c) 3
(d) 4
Solution
y
2
b
ae, –
a
O
x
(0, –b)
Slope of normal at  x1, y1  is
a2 y1
2
b x1
.

b2 
ab2
1
 Slope of normal at  ae,  is 2 or .
a  b ae
e

Solution contd..

b2 
is
 Equation of normal at  ae,


a 

b2e
x  ey  ae 
a
If it passes through (0, –b), then
b2e
be = ae –
or ab = a2 – b2 = a2 e2
a
 a2b2  a4e4


 a2  a2 1  e2  a4e4  e4  e2  1
Solution contd..
Short cut:
As e < 1 for ellipse
e2  e4  2
Now check with options.
e2  e4  1 is the only possibility.
Hence, answer is (a).
Class Exercise - 4
A tangent to the ellipse
x2
a2

y2
b2
1
x2 y2
meets the ellipse

 ab
a
b
in the points P and Q. Prove that the
tangents at P and Q are at right angles.
Solution
Let the tangents at P, Q intersect at
R(h, k). Then according to the given
conditions, chord of contact of
x2 y2

 a  b touches
R w.r.t.
a
b
x2
a2

y2
b2
 1.
hx ky
x2 y2
i.e.

 a  b touches 2  2  1
a
b
a
b
2
2
bh
b
x
y
y
x   a  b  is tangent to
 2 1
2
ak
k
a
b
Solution contd..
 a  b
2
2
2  bh 
2

a

b
 ak 
2
k


b2

c 2  a2m2  b2

2
2
or
a

b

h

k


 a  b b  b h  b k
2 2
2 2
2 2
2
Hence, locus of (h, k) is
x2  y2   a  b   a  a  b   b  a  b 
2
which is the equation of director circle of
x2
y2

1
a a  b b a  b
Hence, tangents at P, Q intersect at right angle.
Class Exercise - 5
Find the locus of mid-points of normal
chords of the ellipse
x2
a
2

y2
b
2
 1.
Solution
Let (h, k) be the mid-point. Then its
2
2
hx
ky
h
k
equation is given by
 2  2 2 .
2
a
b
a
b
If it is normal at
'  ', then this equation is same as
ax sec   by cosec  a2  b2
asec  bcosec a2  b2


 2
h
k
h
k2
 2
2
a2
b2
a
b
Solution contd...
 h2 k 2 
 h2 k 2 
a 2  2 
b  2  2 
a

a
b 
b 


 cos  
, sin  
h 2
k 2
2
2
a

b
a

b
a2
b2




As sin2   cos2   1
2
 h2 k 2 
 2  2 
6
6

a
b
a
b


 2  1

2  2

2
2
h
k


a b


x
y 
Locus of (h, k) is  2  2 
b 
a
2
2
2
 a6 b6 
2
2
 2  2   a  b
y 
x


2
Class Exercise - 6
If P and D be the ends of semi-conjugate
diameters, find the locus of foot of
perpendicular from centre upon PD.
Solution
P   a cos , b sin   ,


 


D   a cos     , b sin        asin , bcos 
2
2 



Equation of PD is
x
 y




cos      sin      cos
a
4 b
4
4


...(i)
[Using equation of chord joining  and  ]
Equation of line perpendicular to above
line passing through (0, 0) is given by
x
 y



sin      cos      0
b
4 a
4


Solution contd..
  by



sin     
cos    
4  ax
4




cos
ax cos


4 
4
From (i) we get cos     
4  x y2

x2  y2

a ax

by cos


4
 sin      2
4  x  y2





sin      cos2      1  a2 x2  b2 y2  2 x2  y2
4
4


2

2
Class Exercise - 7
Find the locus of the middle points of chord
of an ellipse which are drawn through the
positive end of the minor axis.
Solution
Let (a cos  , b sin ) be the coordinate of
the other extremities of the chord of
ellipse
x2
a2

y2
b2
 1 . The positive end of the
minor axis is clearly (0, b).
Let (x, y) be the mid-point of the chord.
x 
a cos   0
2
b  bsin 
y
...(i)
2
2x
2y  b
cos  
, sin  
a
b
...(ii)
Solution contd..
Squaring and adding (i) and (ii),
4x
a
2
2
 2y  b 


 b 
2

4x 2
a
2

4y 2  4yb  b2
b2
On simplifying the required locus of
 x2 y2
y
(x, y) is  2  2  
b 
b
 a
Class Exercise - 8
Prove that the area of the triangle form by
three points on an ellipse, whose eccentric
angles are , , and  is
    
 
2absin 
sin 
 sin 



 2   2 
 2 
Prove also that its area is to the area of the
triangle formed by the corresponding points
on the auxiliary circle as b : a.
Solution
Let the coordinates of the given points
on the ellipse
x2
2
a

y2
2
 1 be acos , bsin ,
b
acos , bsin  be acos , bsin 
Area of the triangle formed by these points

1
2
1 acos  bsin 
1 acos  bsin 
1 acos  bsin 
1
 ab sin      sin      sin     
2
Solution contd..
   

  
 2absin 

sin

sin




 ... (i)
 2 
 2 
 2 
CD
C–D
cos
, etc.
Using sinC  sinD  2sin
2
2
Second Part:
Clearly, coordinates of the points of the auxiliary
circle corresponding to the given three points may
be obtained by putting b = a.



sin
 sin
... (ii)
2
2
2
b
The ratio of the area is .
a
So area of   2a2 sin
Class Exercise - 9
Prove that the straight line
is a normal to the ellipse if
a2
l
2

b2
m2
a


2
b
n2

2 2
.
x  my  n
Solution
The normal at any point P acos ,bsin 
of the ellipse
x2
a2

y2
b2
 1 is given by
ax
by

 a2  b2
cos  sin 
Comparing the given line, x  my  n
and the equation of the normal
ax
by

 a2  b2
cos  sin 
cos 
msin 
n


a
b
a2  b2
Solution contd..
cos  

an
l a2  b2

, sin  
Squaring and adding

an

2
2
cos   sin   
2
2
 a b


i.e.
a2
2

b2
m2
a


2
b
n2
2

2

bn
m a2  b2

2



bn


 
2
2

m a b



2






Class Exercise - 10
The tangents drawn from a point P to
the ellipse make angles 1 and 2 with
the major axis. Find the locus of P
when 1 + 2 is constant = 2.
Solution
Let the equation of the tangent to the
ellipse,
x2
a2

y2
b2
 1 be y  mx  a2m2  b2
and let P(h, k) lies on the tangent.
Then k  mh  a2m2  b2
k  mh
2


 a2m2  b2




m2 h2  a2  2mhk  k2  b2  0
Let tan 1  m1 and tan 2  m2
Solution contd..
m1  m2 
m1m2 
2


2hk
h2  a2

 tan 1  tan 2
k b
h2  a2


t an 1  tan 2 
2
2
h

a


k 2  b2
2

or
By hypothesis 1  2  2.
tan  1  2   tan2 = Constant.
tan 1  tan 2
 tan 2
1  tan 1 tan 2
Solution contd..

2hk
h2  a2

k b 

1
2
2
h

a


2
2
 tan2
Locus is   x2  2xycot 2  y2  a2  b2 


Thank you