Question

Find the equation of the hyperbola satisfying the give 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 $\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1.$
Since the foci are (±4, 0), c = 4.
Length of latus rectum = 12

$⇒ \frac{2b^2}{a} = 12$
$2b^2 = 12a$

$b^2 = \frac{12a}{2 }= 6a$

We know that
$a^2 \+ b^2 = c^2$
$a^2 \+ 6a = 16$
$a^2 \+ 6a – 16 = 0$
$a^2 \+ 8a – 2a – 16 = 0$
$(a + 8) (a – 2) = 0$
$a = -8$ or $2$

Since a is non-negative, $a = 2$.

$∴ b^2 = 6a = 6 \times 2 = 12$

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