Question

In each of the following find the equations of the hyperbola satisfying the given conditions foci (±4, 0), the latus-rectum = 12

Answer 1

Given: Foci (±4, 0), the latus-rectum = 12 

To find: equation of the hyperbola 

Formula used: 

Standard form of the equation of hyperbola is,

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

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

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

According to the question:

We know, 

b² = a²(e² – 1)

⇒ 6a = 16 – a² 

⇒ a² + 6a – 16 = 0 

⇒ a² + 8a – 2a – 16 = 0 

⇒ a(a + 8) – 2(a + 8) = 0

⇒ (a + 8)(a – 2) = 0 

⇒ a = -8 or a = 2 

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

⇒ a = 2 

⇒ a² = 4 b² = 6a 

⇒ b² = 6(2) 

⇒ b² = 12 

Hence, equation of hyperbola is: