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Solved Examples on Hyperbola

Illustration 1: If e1 is the eccentricity of the ellipse x2/16 + y2/25 = 1 and e2 is the eccentricity of the hyperbola passing through the foci of the ellipse and e1e2 = 1, then find the equation of the hyperbola. (2006)  

Solution: The eccentricity of x2/16 + y2/25 = 1 is

e1 = √(1-16/25) = 3/5

e2 = 5/3

This is obtained using the relation e1e2 = 1.

Hence, the foci of the ellipse are (0, ± 3)

Hence, the equation of the hyperbola is x2/16 - y2/9 = -1.

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Illustration 2: A hyperbola, having the transverse axis of length 2 sin θ, is confocal with the ellipse 3x2 + 4y2 = 12. Then its equation is (2007)

1. x2cosec2 θ – y2sec2 θ = 1                    2. x2sec2 θ – y2cosec2θ = 1

3. x2sin2 θ – y2cos2 θ = 1                        4. x2cos2 θ – y2sin2 θ = 1

Solution: The given ellipse is x2/4 + y2/3 = 1

This gives a = 2 and b = √3

Hence, 3 = 4(1-e2) which gives e = 1/2

So, ae = 2.1/2 = 1

Hence, the eccentricity e1 for the hyperbola is given by

1 = e1 sin θ which means e1 = cosec θ.

So, b2 = sin2 θ(cosec2 θ – 1) = cos2 θ

Hence, the equation of hyperbola is x2/sin2 θ - y2/cos2θ = 1

Or x2cosec2 θ – y2sec2 θ = 1                   

Illustration 3: The circle x2 + y2 – 8x = 0 and the hyperbola x2/9 - y2/4 = 1 intersect at the points A and B. (2010)

1. Equation of a common tangent with positive slope to the circle as well as to the hyperbola is ….?

2. Also find the equation of the circle with AB as its diameter.

Solution: (1) Equation of tangent to hyperbola having slope m is y = mx + √9m2-4

Equation of tangent to circle is y = m(x-4) + √16m2 +16

These two equations will be identical for m = 2/√5...

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