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.

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       

Hence, the equation of common tangent is 2x - √5y + 4 = 0

(2) The equation of the hyperbola is x²/9 - y²/4 = 1 and that of circle is x² + y² – 8x = 0.

For their points of intersection x²/9 + (x²– 8x)/ 4 = 1

So, this gives 4x² + 9x² -72x = 36

So, 13x² -72x -36 = 0   

This gives x = 6 and -13/6

But x = -13/6 is not acceptable

Now, x = 6, y = ± 2√3

Required equation is (x-6)² + (y+2√3)(y-2√3) = 0

This gives x² + y² -12x + 24 = 0