Question

The vertices of a hyperbola are at $(0,0)$ and $(10,0)$ and one of its foci is at $(18,0)$. The possible equation of the hyperbola is -

• A. $\frac{x^2}{25}-\frac{y^2}{144}=1$
• B. $\frac{(x-5)^2}{25}-\frac{y^2}{144}=1$
• C. $\frac{x^2}{25}-\frac{(y-5)^2}{144}=1$
• D. $\frac{(x-5)^2}{25}-\frac{(y-5)^2}{144}=1$

Answer 1

Answer: (B)

Centre of hyperbola is $(5,0)$, so equation is
$\frac{(x-5)^2}{a^2}-\frac{y^2}{b^2}=1$
$a=5, a e-a=8 \Rightarrow e=\frac{13}{5}$
$b^2=144$
So equation is $\frac{(x-5)^2}{25}-\frac{y^2}{144}=1$.