Question

Find the equation of the hyperbola, referred to its axes as axes of coordinates, given that the distance of one of its vertices from the foci are $9$ and $1$ units, is $\frac{x^2}{16}-\frac{y^2}{9}=1$.

Answer 1

Let the equation of hyperbola be $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ ...(1)

Its vertices are $A(a,0)$ and $A^{\prime }(-a,0)$ and foci are $S(ae,0)$ and $S^{\prime }(-ae,0)$.

Given $:S^{\prime }A=9$ and $SA=1$

$\Rightarrow a+ae=9$ and $ae-a=1$

$\Rightarrow a(1+e)=9$ and $a(e-1)=1$

$\Rightarrow \frac{a(1+e)}{a(e-1)}=\frac{9}{1}\Rightarrow 1+e=9e-9\Rightarrow e=\frac{5}{4}$

$\because a(1+e)=9\therefore a(1+\frac{5}{4})=9\Rightarrow a=4$,$b=3$

Thus, from (1), equation of hyperbola is $\frac{x^2}{4^2}-\frac{y^2}{3^2}=1$