Question

Find the equation of the hyperbola with foci at (±6, 0) and eccentricity 3/2 .

Answer 1

Answer:

The answer is $25x^2-16y^2=400$

Step-by-step explanation:

Let $a$ be the length semi transverse axis and $b$ be the length of its semi conjugate axis

Given foci are $(\pm 6,0)$ and eccentricity, $e=\frac{3}{2}$

$\Rightarrow ae=6\Rightarrow a(\frac{3}{2})=6\Rightarrow a=4$

Using $b^2=a^2(e^2-1)=4(\frac{9}{4}-1)=5$

Hence the hyperbola is

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

             $\Rightarrow \frac{x^2}{16}-\frac{y^2}{25}=1\\\Rightarrow 25x^2-16y^2=400$