Question

Find the equation of the hyperbola satisfying the give conditions: Foci ($± 3\sqrt5, 0$), the latus rectum is of length 8.

Answer 1

Foci ($± 3\sqrt5, 0$), the latus rectum is of length 8.
Here, the foci are on the x-axis.
Therefore, the equation of the hyperbola is of the form $\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1.$
Since the foci are ($± 3\sqrt5, 0$), $c = ± 3\sqrt5 .$
Length of latus rectum = 8
$⇒\frac{ 2b^2}{a} = 8$
$⇒ 2b^2 = 8a$
$⇒ b^2 = \frac{8a}{2}$
$= 4a$

We know that
a2 + b2 = c2
a2 + 4a = 45
a2 + 4a – 45 = 0
a2 + 9a – 5a – 45 = 0
(a + 9) (a -5) = 0
a = -9 or 5

Since a is non-negative, a = 5.

$∴ b^2 = 4a= 4 \times 5= 20$
Thus, the equation of the hyperbola is $\frac{x^2}{25} –\frac{ y^2}{20} = 1$