Question

The vertices of a hyperbola are at $(0,0)$ and $(10,0)$ and one of its foci is at $(18,0)$ . The 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 $\frac{(x-5{)}^2}{25}-\frac{y^2}{144}=1$

vertices are $A(0,0)$ and $A^{\prime }(10,0)$

$A{A}^{\prime }=2a$

$\Rightarrow 10=2a$

$\Rightarrow a=5$

Focus is

$ae-a=8$

$\Rightarrow e-1=\frac{8}{5}$

$\Rightarrow e=1+\frac{8}{5}=\frac{13}{5}$

Now, $b=a\sqrt{e^2-1}$

$b=5\sqrt{\frac{169}{25}-1}$

$=5\frac{12}{5}=12$

Since, centre $=(ae,0)=(5,0)$

$\therefore$ Required equation

$\frac{(x-5{)}^2}{25}-\frac{y^2}{144}-1$

Hence proved