HYPERBOLA-2 (Conjugate Hyperbola)

Position of a point

Let a point P(x1,y1) and equation of hyperbola be x2ya2y2 b21=0

S1=x12a2y12 b21

If S1>0, point is inside the hyperbola.

If S1=0, point is on the hyperbola.

If S1<0, point is outside the hyperbola.

Conjugate Hyperbola

Corresponding to every hyperbola there exists a hyperbola such that the transverse axis and conjugate axis of one is equal to the conjugate axis and transverse axis of the other. Such hyperbolas are known as conjugate to each other.

Therefore for the hyperbola x2a2y2 b2=1

Conjugate hyperbola is x2a2y2 b2=1

Let e1 be the eccentricity of x2a2y2 b2=1 and e2 be

the eccentricity of x2a2y2 b2=1 then e12=1+b2a2=a2+b2a2

and e22=1+a2 b2=b2+a2 b2

1e12+1e22=1

The foci of a hyperbola and its conjugate hyperbola are concyclic and form the vertices of a square.

Auxiliary circle and eccentric angle

A circle drawn with centre O and transverse axis as diameter is known as auxiliary circle. Equation of auxiliary circle is x2+y2=a2

A is any point on the circle whose coordinates are (acosθ,asinθ), where θ is known as eccentric angle. Now, In OAB,OA=a

cosθ=aOBOB=asecθ

P point lies on hyperbola, so a2sec2θa2y2 b2=1y=±btanθ

P(asecθ,btanθ)0θ<2π

The equations x=asecθ and y=btanθ represents a hyperbola. So, the parametric form of the hyperbola x2a2y2b2=1 can be represented as x=asecθ,y=btanθ.

For the hyperbola (xh)2a2(yk)2b2=1, parametric form is x=h+asecθ;y=k+btanθ.

Hyperbola Conjugate Hyperbola
Equation x2a2y2 b2=1 x2a2+y2 b2=1
Centre (0,0) (0,0)
Vertice (a,0)&(a,0) (0, b)&(0,b)
Foci (ae,0)&(ae,0) (o,be)&(0,be)
Length of transverse axis 2a 2 b
Length of conjugate axis 2 b 2a
Length of latus rectum 2 b2a 2a2 b
Equation of transverse axis y=0 x=0
Equation of conjugate axis x=0 y=0
Equation of directrices x=±ae y=±be
Eccentricity e=1+b2a2 or b2=a2(e21) e=1+a2 b2 or a2=b2(e21)

Examples

1. The point (at2+aat,bt2b2t) lies on the, [ for all values of t(t0)]

(a) circle

(b) parabola

(c) ellipse

(d) hyperbola

Show Answer

Solution:

Let x=at2+a2t and y=bt2b2t

2xa=t+1t and 2yb=t1t

Squaring and subtracting, we get

4x2a24y2 b2=4x2a2y2 b2=1

Answer: d

2. The position of the point (5,4) relative to the hyperbola 9x2y2=1 is

(a) on the hyperbola

(b) outside the hyperbola

(c) Inside the hyperbola

(d) can not say

Show Answer

Solution:

S=9x2y21S1=9(5)2(4)21=225161=208>0

point (5,4) lies inside the hyperbola.

Answer: c.

3. Two circles are given such that they neither intersect nor touch. The locus of centre of variable circle which touches both the circles externally is

(a) a circle

(b) a parabola

(c) an ellipse

(d) a hyperbola

Show Answer

Solution:

Let radii of the fixed circles be r1 and r2 and radius of variable circle be r

Let variable circles with centre C touches two fixed circles with centre C1 and C2.

Then CC1=r+r1 and CC2=r+r2CC1CC2=r1r2= constant

Locus of C is hyperbola whose foci are C1 and C2

Answer: d

4: If the latus rectum subtends a right angle at the centre of the hyperbola x2a2y2b2=1, then its eccentricity is

(a) 5+12

(b) 3

(c) 52

(d) 5+32

Show Answer

Solution:

mAC×mBC=1

b2 a.ae ×b2 a.ae =1

b4=a4e2

b4a4=e2

(e21)2=e2

e43e2+1=0

e2=3±52e2=3+52e=3+52

[if e=352<1 but ecentricity of hyperbola >1 so neglecting this value of e]

Answer: d

Practice questions

1. The equation of hyperbola whose foci are (8,3),(0,3) and eccentricity is 43 is

(a). (x4)29(y3)27=1

(b). (x4)27(y3)29=1

(c). (x3)29(y4)27=1

(d). none of these

Show Answer Answer: (a)

2. If S and S be the foci, C the centre and P be a point on a rectangular hyperbola then SP×SP is equal to

(a). 2.SP

(b). (SP)2

(c). (CP)2

(d). 2.CP

Show Answer Answer: (c)

3. If e and e be the eccentricity of a hyperbola and its conjugate, then

(a). 1e21e2=1

(b). e2+e2=1

(c). e2e2=1

(d). 1e2+1e2=1

Show Answer Answer: (d)

4. The foci of a hyperbola coincide with the foci of the ellipse x225+y29=1, The equation of hyperbola if its eccentricity is 2 is

(a). 3x2y2=12

(b). 4x2y2=12

(c). x23y2=12

(d). x24y2=12

Show Answer Answer: (a)

5. The equation (x4)2+(y2)2+(x+4)2+(y2)2=8 represents

(a). an ellipse

(b). a parabola

(c). a pair of coincident line segment

(d). hyperbola

Show Answer Answer: (c)

6. For hyperbola x2cos2αy2sin2α=1, which of the following remains constant with change in α.

(a). abscissae of vertices

(b). abscissae of foci

(c). eccentricity

(d). directrix

Show Answer Answer: (b)

7. Two rods are rotating about two fixed points in opposite directions. If they start from their position of co-incidence and one rotates at the rate double that of the other, then locus of point of intersections of two rods is

(a). a parabola

(b). a circle

(c). an ellipse

(d). a hyperbola

Show Answer Answer: (d)

8. The equations x21ry21+r=1,r>1 represents

(a). an ellipse

(b). a circle

(c). a hyperbola

(d). None of these

Show Answer Answer: (d)

9. The equation 2x2+3y28x18y+35=k represents

(a). no locus if k>0

(c). a point if k=0

(b). an ellipse if k<0

(d). a hyperbola if k>0

Show Answer Answer: (c)

10. A hyperbola having the transverse axis of length 2sinθ is confocal with the ellipse 3x2+4y2=12. Then its equation is

(a). x2cosec2θy2sec2θ=1

(b). x2sec2θy2cosec2θ=1

(c). x2sin2θy2cos2θ=1

(d). x2cos2θy2sin2θ=1

Show Answer Answer: (a)