Question

The equation of the hyperbola whose foci are $(6,5),(-4,5)$ and eccentricity $5/4$ is?

• A. $\frac{(x-1{)}^2}{16}-\frac{(y-5{)}^2}{9}=1$
• B. $\frac{x^2}{16}-\frac{y^2}{9}=1$
• C. $\frac{(x-1{)}^2}{16}-\frac{(y-5{)}^2}{9}=-1$
• D. $\frac{(x-1{)}^2}{4}-\frac{(y-5{)}^2}{9}=1$

Answer 1

The correct option is A $\frac{(x-1{)}^2}{16}-\frac{(y-5{)}^2}{9}=1$

Let the centre of hyperbola be $(\alpha ,\beta )$

As $y=5$ line has the foci, it also has the major axis.

$\therefore \frac{(x-\alpha {)}^2}{a^2}-\frac{(y-\beta {)}^2}{b^2}=1$

Midpoint of foci = centre of hyperbola

$\therefore \alpha =1,\beta =5$

Given, $e=\frac{5}{4}$.

We know that foci is given by $(\alpha \pm ae,\beta )$

$\therefore \alpha +ae=6$

$\Rightarrow 1+\frac{5}{4}a=6\Rightarrow a=4$

Using $b^2=a^2(e^2-1)$

$\Rightarrow b^2=16(\frac{25}{16}-1)=9$

$\therefore Equationofhyperbola\Rightarrow \frac{(x-1{)}^2}{16}-\frac{(y-5{)}^2}{9}=1$