Question

Find the equation of the hyperbola satisfying the given condition,

Vertices $(0,\pm 3),foci(0,\pm 5)$

Answer 1

The vertices are $(0,\pm 30$ which lie on y - axis.
So the equation of hyperbola in standard form
is, $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1\therefore \text{The vertices}(0,\pm a)\text{is}(0,\pm 3)\Rightarrow a=3\text{foci}(0,\pm ae)\text{is}(0,\pm 5)\Rightarrow ae=5\text{Now}ae=5\Rightarrow e=\frac{5}{a}\Rightarrow e=\frac{5}{3}\text{We know that}b=a\sqrt{e^2-1}=3\sqrt{\frac{25}{9}-1}=\frac{3\times 4}{3}=4$
Thus required equation of hyperbola is,
$\frac{y^2}{(3{)}^2}-\frac{x^2}{(4{)}^2}=1\Rightarrow \frac{y^2}{9}-\frac{x^2}{16}=1.$