Question

Find the equation of the hyperbola satisfying the given conditions:

Foci (±4, 0), the latus rectum is of length 12.

Answer 1

Foci (±4, 0), the latus rectum is of length 12.

Here, the foci are on the x-axis.

Therefore, the equation of the hyperbola is of the form ${\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2}=1}$.

Since the foci are ${\displaystyle (\pm 4,0)}$, C = 4

Length of latus retum = 12

${\displaystyle \frac{2b^2}{a}=12}$

= b² = 6a

We know that a² + b² = c²

∴ a² + 6a = 16

= a² + 6a - 16 = 0

= a² + 8a - 2a - 16 = 0

= (a + 8) (a - 2) = 0

= a = -8, 2

but a ≠ –8,

∴ a = 2, a² = 4

∴ b² = 6a = 6 × 2 = 12

Thus, the equation of the hyperbola is ${\displaystyle \frac{x^2}{4}-\frac{y^2}{12}=1}$.