Question

In each of the following find the equations of the hyperbola satisfying the given conditions foci (0, ± 12), latus-rectum = 36.

Answer 1

Given: Foci (0, ±12), the latus-rectum = 36 

To find: equation of the hyperbola 

Formula used: 

The standard form of the equation of the hyperbola is,

\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = -1\)

Coordinates of the foci for a standard hyperbola is given by (0, ±be) 

Length of latus rectum is \(\frac{2a^2}{b}\)

According to the question:

We know, 

a² = b²(e² – 1)

⇒ 18b = 144 – b² 

⇒ b² + 18b – 144 = 0 

⇒ b² + 24b – 6b – 144 = 0 

⇒ b(b + 24) – 6(b + 24) = 0

⇒ (b + 24)(b – 6) = 0 

⇒ b = -24 or b = 6 

Since b is a distance, and it can’t be negative 

⇒ b = 6 

⇒ b² = 36 a² = 18b 

⇒ a² = 18(6) 

⇒ b² = 108 

Hence, equation of hyperbola is: