Yimin Math Centre
4 Unit Math Homework for Year 12
Student Name:
Grade:
Date:
Score:
Table of contents
4
Topic 2 — Conics (Part 4)
1
4.1
Further Geometric Properties of the Ellipse and Hyperbola . . . . . . . . . . . . . . .
1
4.2
The Rectangular Hyperbola . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
4.2.1
Cartesian Equation of the Rectangular Hyperbola . . . . . . . . . . . . . . . .
3
4.2.2
Parametric Equations of the Rectangular Hyperbola . . . . . . . . . . . . . . .
3
4.2.3
Equation of a Chord of the Hyperbola xy = c2 . . . . . . . . . . . . . . . . .
5
4.2.4
Equations of Tangents and Normals to the hyperbola xy = c2
. . . . . . . . .
6
4.2.5
Equation of the Chord of Contact of Tangent from an External Point . . . . . .
8
4.2.6
Further geometric properties of the rectangular hyperbola . . . . . . . . . . . .
9
4.2.7
Locus Problem and the Rectangular Hyperbola . . . . . . . . . . . . . . . . .
11
This edition was printed on June 17, 2013.
Camera ready copy was prepared with the LATEX2e typesetting system.
Copyright © 2000 - 2013 Yimin Math Centre (www.yiminmathcentre.com)
4 Unit Math Homework for Year 12
Year 12 Topic 4 Homework
4
4.1
Page 1 of 13
Topic 2 — Conics (Part 4)
Further Geometric Properties of the Ellipse and Hyperbola
Definition:
1. The segment of the tangent to an ellipse or hyperbola between the point of the contact
and the directrix subtends a right angle at the corresponding focus. (P S ⊥ ST )
2. The tangent at a point P on the ellipse or hyperbola is equally inclines to the focal
chords through P. (∠SP T = ∠S 0 P T 0 )
3. The normal at a point P on the ellipse or hyperbola is equally inclined to the focal
chords through P.
Copyright © 2000 - 2012 Yimin Math Centre (www.yiminmathcentre.com)
Year 12 Topic 4 Homework
Page 2 of 13
2
2
Example 4.1.1 P (a sec θ, b tan θ) lies on the hyperbola xa2 − yb2 = 1. The tangent at P meets the
tangents at the ends of the major axis at Q and R. Show that QR subtends a right angle at either
2
focus. Deduce that if P is the point 5, 34 on the hyperbola x9 − y 2 = 1, with foci S and S’, then Q,
S, S, S’ are concyclic and find the equation of the circle through these points.
Solution: Let the tangent at P meet x = a, x = −a in the Q, R respectively.
Let QR meet the y-axis in C.
θ
θ
Tangent PR has equation x sec
− y tan
= 1.
a
b
b
Hence Q has coordinates a, tan θ (sec θ − 1) and R has coordinates −a, − tanb θ (sec θ + 1)
θ−1
Gradient QS.gradient RS = a b(sec
× −a−b(secθ+1
tan θ(1−e)
tan θ(1+e)
2
2
(sec θ−1)
b
.
= a2 (1−e
2) ×
tan2 θ
2
2 2
Then b = a (e − 1) ⇒ gradient QS× gradient RS = −1, ∴ QS ⊥ RS
Similarly, replacing e by −e, QS 0 ⊥ RS 0 .
Hence QR subtends angles of 90◦ at each of S and S’, and Q, S, R, S’ are noncyclic,
with QR the diameter of the circle through the points.
The y-axis is the perpendicular bisector of the chord SS’,
Hence the centre of this circle is the point C where the diameter QR meets the y-axis.
2
If P (5, 34 lies in the hyperbola x9 − y = 1, then QR has equation 5x
− 4y
= 1 and meets
9
3
3
the y-axis in C 0, 4 .
√
Also b2 = a2 (e2 − 1) gives e = 10
10, 0).
,
and
S
ahs
coordinates
(
9
2
169
2
0
Hence CS = 16 and the circle through Q, R, S, S has equation x2 + y + 43 = 169
.
16
Copyright © 2000 - 2012 Yimin Math Centre (www.yiminmathcentre.com)
Year 12 Topic 4 Homework
4.2
4.2.1
Page 3 of 13
The Rectangular Hyperbola
Cartesian Equation of the Rectangular Hyperbola
Definition: A hyperbola is called rectangular if its asymptotes meet at right angles.
2
2
The asymptotes of xa2 − yb2 = 1 have equations y = ± ab x.
Hence asymptotes perpendicular ⇒ ( ab ) × (− ab = −1), ⇒ a = b.
q
√
2
Then b2 = a2 (e2 − 1), ⇒ e = (1 + ab 2 ), ⇒ ∴ e = 2.
Hence a rectangular hyperbola, with major axis along the x-axis,
has equation x2 − y 2 = a2 ,
√
eccentricity e = 2 and asymptotes y = ±x.
4.2.2
Parametric Equations of the Rectangular Hyperbola
Definition: The standard parametric from of the rectangular hyperbola xy = c2 is
x = ct and y = ct ,
where the value of parameter t depends on the position of P (ct, ct ) on the curve
as in the figure shown below:
Copyright © 2000 - 2012 Yimin Math Centre (www.yiminmathcentre.com)
Year 12 Topic 4 Homework
Exercise 4.2.1
1. For the rectangular hyperbola xy = 4, find the parametric equation.
2. For the rectangular hyperbola x = 4t, y = 4t , find the Cartesian equation.
3. Find the parametric equation of the rectangular hyperbola xy = 25.
4. Find the Cartesian equation of the rectangular hyperbola x = 3t, y = 3t .
Copyright © 2000 - 2012 Yimin Math Centre (www.yiminmathcentre.com)
Page 4 of 13
Year 12 Topic 4 Homework
4.2.3
Page 5 of 13
Equation of a Chord of the Hyperbola xy = c2
Cartesian form:
Let P (x1 , y1 ), Q(x2 , y2 ) lie on the hyperbolaxy = c2 .
2
2
−y1
1
c2
= x2 −x
− xc 2 = x−c
The gradient of PQ is xy22 −x
x2
1
1
1 x2
1
∴ chord PQ has equation c2 x + x1 x2 y = k, k is constant.
∴ k = c2 (x1 + x2 ).
The chord from P (x1 , y1 ) to Q(x2 , y2 ) on xy = c2
has equation c2 x + x1 x2 y = c2 (x1 + x2 ).
Parametric form:
c
c
Let P cp, p , Q cq, q lie on the hyperbola xy = c2 .
c( 1 − 1 )
q
p
1
= − pq
The gradient of PQ is c(q−p)
∴ chord PQ has equation x + pqy = k, k is constant.
P (cp, pc ) lies on P Q ⇒ ∴ k = c(p + q).
The chord from P (cp, pc ) to Q(cq, qc ) on xy = c2
has equation x + pqy = c(p + q).
Example 4.2.1 The points P (cp, pc ) and Q(cq, qc ) lie on the rectangula hyperbola xy = c2 . The
chord PQ subtends a right angle at the another point R(cr, rc ) on the hyperbola. Show the the
normal at R is parallel to PQ.
Solution: The gradient of PR is
c( p1 − r1 )
c(p−r)
c( 1q − r1 )
1
= − pr
,
the gradient of QR is c(q−r) = − qr1 .
Therefore P Q ⊥ QR, ⇒ gradient P R× gradient QR = −1,
1
1
2
∴ pqr
2 = −1 ⇒ r = − pq .
The normal at the point R(cr, rc ) has gradient of r2 ,
c( 1 − 1
1
the gradient of PQ is c(Pp −q)q = − pq
.
1
Since r2 = − pq
, then gradient of the normal at R equal to gradient of PQ.
Thus the normal at the point R is parallel to the chord PR.
Copyright © 2000 - 2012 Yimin Math Centre (www.yiminmathcentre.com)
Year 12 Topic 4 Homework
4.2.4
Page 6 of 13
Equations of Tangents and Normals to the hyperbola xy = c2
Cartesian from of equation of tangent:
dy
=0
By implicit differentiation, xy = c2 , ⇒ y + x dx
dy
y
∴ dx = − x
Gradient of tangent at P (x1 , y1 ) on the hyperbola is − xy11
∴ tangent at P has equation y1 x + x1 y = k, k is constant.
By P lies on tangent ∴ k = 2x1 y1 = 2c2 .
The tangent to xy = c2 atP (x1 , y1 ) has equation xy1 + yx1 = 2c2
Cartesian from of equation of normal:
Also the normal at P (x1 , y1 ) and equation y − y1 = xy11 (x − x1 ).
The normal to xy = c2 at P (x1 , y1 ) has equation xx1 − yy1 = x21 − y 2 .
Parametric form of equation of tangent:
= c, and y = ct , ⇒ dy
= − tc2 ,
x = ct, ⇒ dx
dt
dt
dy
∴ dx
= − t12
Gradient of tangent at P (cp, pc ) on the hyperbola is − p12 ,
∴ tangent at P has equation x + p2 y = k, k is constant.
But P lies on tangent ∴ k = 2cp. 2
The tangent to xy = c at P cp, pc has equation x + p2 y = 2cp.
Parametric form of equation
of normal:
Also the normal at P cp, pc has gradient P 2 and equation y − Pc = p2 (x − cp).
1
1
c
2
2
The normal to xy = c at P cp, p has equation px − p y = c p − p2 .
Example 4.2.2 Find the equation of the tangent and the normal to the rectangular hyperbola
x = 2t, y = 2t at the point where t = 4.
Solution: For the hyperbola x = 2t, y = 2t , where we have c = 2.
Hence the tangent to the hyperbola x = 2t, y = 2t at point where t = 4
has equation x + t2 y = 2ct, ⇒ x + 16y = 16
and the normal has equation tx − 1t y = c t2 − t12 ⇒ 32x − 2y = 255.
Copyright © 2000 - 2012 Yimin Math Centre (www.yiminmathcentre.com)
Year 12 Topic 4 Homework
Page 7 of 13
Exercise 4.2.2
1. Find the equations of the tangent and the normal to the rectangular hyperbola xy = 12 at the
point (-3, -4).
2. Find the equations of the tangent and the normal to the rectangular hyperbola x = 3t, y =
the point t = −1.
3
t
at
3. Find the equation of chord of contact of tangent from the point (1, -2) ro the hyperbola xy = 6.
4. Find the equation of the tangent and the noram to the rectangular hyperbola xy = 8 at the point
(4, 2).
Copyright © 2000 - 2012 Yimin Math Centre (www.yiminmathcentre.com)
Year 12 Topic 4 Homework
4.2.5
Page 8 of 13
Equation of the Chord of Contact of Tangent from an External Point
Definition: T lies on the tangent PT and QT, ∴ x0 y1 + y0 x1 = 2x2 and x0 y 2 + y0 x2 = 2c2 .
Hence P (x1 , y1 ) and Q(x2 , y2 ) both satisfy x0 y + y0 x = 2c2 .
∴ this linear equation must be the equation of PQ.
The chord contact of tangent from an external point T (x0 , y0 ) to
the hyperbola xy = c has equation xy0 + yx0 = 2c2 .
Example 4.2.3 Find the equation of the chord of contact of tangent from the point (2, 1) to
xy = 10.
Solution: For the hyperbola xy = 10, we have c2 = 10.
Hence the chord of contact of tangents from the point T (x0 , y0 ) = T (2, 1)
to the hyperbola xy = 10 has equation xy0 + yx0 = 2c2 ⇒ x + 2y = 20.
Exercise 4.2.3 Find the equation of the chord of contact of tangents from the point (−1, −3) to
the rectangular hyperbola xy = 4. Hence find the coordinated of their points of contact and the
equations of these tangents.
Copyright © 2000 - 2012 Yimin Math Centre (www.yiminmathcentre.com)
Year 12 Topic 4 Homework
4.2.6
Page 9 of 13
Further geometric properties of the rectangular hyperbola
Definition:
1. The area of the triangle bounded by a tangent and the asymptotes is a constant.
Let the tangent at P (ct, ct on xy = c2 meet the x-axes and y-axes in R and T respectively.
The tangent has equation x + t2 y = 2ct. At T, x = 0, y = 2ct .
∴ Area 4OT R = 12 × 2ct × 2ct = 2c2 .
2. The length of the intercept cut off a tangent by the asymptotes is twice the distance
from the point of the intersection of the asymptotes to the point of the contact of the
tangent.
2
T R2 = ( 2ct ) + (2ct)= 4c2 (t2 + t12 ), and OP 2 (ct)2 + ( ct )2 = c2 (t2 + t12 ).
T R2 = 4OP 2 , ⇒ ∴ T R = 2OP.
3. The product of the focal distance of a point P on a rectangular hyperbola is equal to
the square of the distance from P to the point of intersection of the asymptotes.
Let xy = c2 have foci S and S’, and directrices m and m’. Let M, M’ be the feet of
the perpendiculars from P (ct, ct to m, m’ respectively.
√
Then P S × P S 0 = eP M × eP√M 0 = 2P M√× P M 0 (since e = 2) ,
√ 2
|ct+ ct + 2c|
|ct+ ct − 2c|
c 2
√
√
∴ P S × P S0 = 2 ×
×
=
|(ct
+
)
|
−
(
2c)
t
2
2
∴ P S × P S 0 = (ct)2 + ( ct )2 = OP 2 .
Copyright © 2000 - 2012 Yimin Math Centre (www.yiminmathcentre.com)
Year 12 Topic 4 Homework
Page 10 of 13
Example 4.2.4 For the rectangular hyperbola xy = 16, find (a) the eccentricity; (b) the coordinates if the foci; (c) the equation s of the directrices; (d) the equations of the asymptotes. Sketch
the hyperbola.
Solution: For the hyperbola xy = 16 we have c2 = 16 ⇒ c = 4.
√
Hence the hyperbola xy = 16 has eccentricity e = 2
√
√
√
√
Foci: S(c 2, c 2), ⇒ S(4 2, 4 2)
√
√
√
√
and S 0 (−c 2, −c 2), ⇒ S 0 (−4 2, −4 2),
√
√
Dirctires: x + y = ±c 2, ⇒ x + y = ±4 2,
Asymptotes: x = 0 and y = 0.
Exercise 4.2.4 The point P (ct, ct ), where t = -1 lies on the rectangular hyperbola xy = c2 . The
tangent and normal at P meet the line y = x at T and N respectively. Show that OT × ON = 4c2 .
Copyright © 2000 - 2012 Yimin Math Centre (www.yiminmathcentre.com)
Year 12 Topic 4 Homework
4.2.7
Page 11 of 13
Locus Problem and the Rectangular Hyperbola
Example 4.2.5 P (ct, ct ) is a point on the rectangular hyperbola xy = c2 . N is the foot of the
perpendicular from P to the x-axis and M is the midpoint of P N . Find the Cartesian equation of
the locus of M as the position of P varies. and describe this locus geometrically.
Solution: M has coordinates (ct, 2tc ).
Hence the locus of M has parametric equations x = ct,
and y = 2tc and Cartesian equation xy = 21 c2 .
M traces a rectangular hyperbola as the position of P varies.
Example 4.2.6 P (3p, p3 ) and Q(3q, 3q ) are point on different branches of the hyperbola xy = 9.
Find the coordinates of the point of intersection T of the tangents at P and Q. Find the locus of T
if the positions of P and Q very so that the chord P Q passes through (0, 4).
(
x + p2 y = 6p (1)
x + q 2 y = 6q (2)
6
(1) − (2) ⇒ (p2 − q 2 )y = 6(p − q), p 6= q ⇒ ∴ y = p+q
6pq
6pq
6
Then (1) ⇒ x = p+q
, ⇒ T has coordinates p+q
, p+q
.
If (0, 4) lies on chord PQ, with equation
x + pqy = 3(p + q),
9
then 4pq = 3(p + q) and T is T 29 , 2pq
.
P, Q on different branches ⇒ pq < 0, ⇒ ∴ locus of T is x = 92 , y < 0
Solution: T lies on the tangent at P and Q. ⇒ At T,
Copyright © 2000 - 2012 Yimin Math Centre (www.yiminmathcentre.com)
Year 12 Topic 4 Homework
Page 12 of 13
Exercise 4.2.5
1. P and Q are variable points on the rectangular hyperbola xy = 9. The tangents at P and Q meet
at R. If P Q passes through the point (6, 2). find the equation of the locus of R.
2. The point P (ct, ct ) lies on the rectangular hyperbola xy = c2 . The tangent at P cuts the x-axis at
X and the y-axis at Y. Show that the area of 4Y OX is independent of t.
3. On the rectangular hyperbola xy = c2 there are variable points P and Q. The tangents at P and
Q meet at R. Find the equation of the locus of R if PQ passes through the point (a, 0).
4. The point P (ct, ct ) lies on the rectangular hyperbola xy = c2 . The tangent at P cuts the x-axis at
X and the y-axis at Y. Show that PX = PY.
Copyright © 2000 - 2012 Yimin Math Centre (www.yiminmathcentre.com)
Year 12 Topic 4 Homework
Page 13 of 13
Exercise 4.2.6
1. The point P (ct, ct ) lies on the rectangular hyperbola xy = c2 . Show that the normal at P cuts the
hyperbola again at the point Q with coordinates (− tc3 , −ct3 ). Hence find the coordinates of the
point R where the normal at Q cuts the hyperbola again.
2. The point P (ct, ct ) lies on the rectangular hyperbola xy = c2 . The normal at P meets the x-axis
at A and the tangent at P meets the y-axis at B. M is the midpoint of AB. Find the equation of the
locus of M as P moves on the hyperbola.
Copyright © 2000 - 2012 Yimin Math Centre (www.yiminmathcentre.com)