Question

Find the equation of the hyperbola satisfying the given condition,

Foci $(\pm 5,0)$, the transverse axis is of length 8.

Answer 1

Here foci are $(\pm 5,0)$ which lie on x axis.
So the equation of the hyperbola in standard form is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\therefore \text{foci}(\pm ae,0)\text{is}(\pm 5,0)\Rightarrow ae=5\text{Length of transverse axis}2a=8\Rightarrow a=4\text{Now, ae = 5}\Rightarrow e=\frac{5}{a}\Rightarrow e=\frac{5}{4}\text{We know that}b=a\sqrt{e^2-1}\Rightarrow b=4\sqrt{\frac{25}{16}-1}=4\times \frac{3}{4}=3$
Thus required equation of hyperbola is,
$\frac{x^2}{(4{)}^2}-\frac{y^2}{(3{)}^3}=1\Rightarrow \frac{x^2}{16}-\frac{y^2}{9}=1.$