Question

The vertices of a hyperbola are at $(0,0)$ and $(10,0)$ and one of its focus 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

The correct option is B $\frac{(x-5{)}^2}{25}-\frac{y^2}{144}=1$

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