Question

Find the equation of the hyperbola satisfying the give conditions: Vertices (±7, 0), $e = \frac{4}{3}$

Answer 1

Vertices (±7, 0), $e =\frac{ 4}{3}$
Here, the vertices 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 vertices are (±7, 0), a = 7.

It is given that $e =\frac{ 4}{3}$

$∴ \frac{c}{a} = \frac{4}{3} [e=\frac{c}{a}]$

$⇒ 3c = 4a$
$⇒ 3c = 4(7)$

$⇒ c = \frac{28}{3}$

We know that $a^2 \+ b^2 = c^2$

$7^2 \+ b^2 = (\frac{28}{3})^2$

$b^2 = \frac{784}{9} – 49$

$=\frac{ (784 – 441)}{9}$

$= \frac{343}{9}$

Thus, the equation of the hyperbola is $\frac{x^2}{49} – \frac{9y^2}{343} = 1$