Question

Equation of the hyperbola whose vertices are at ($\pm 3,0$) and focii at ($\pm 5,0$) is

• A. $16x^2-9y^2=144$
• B. $9x^2-16y^2=144$
• C. $25x^2-9y^2=255$
• D. $9x^2-25y^2=81$

Answer 1

The correct option is A $16x^2-9y^2=144$

Given $a=3$ and $ae=5\Rightarrow e=\frac{5}{3}$
Using $e^2=1+\frac{b^2}{a^2}$
$\frac{25}{9}=1+\frac{b^2}{9}\Rightarrow b^2=16$
Therefore required hyperbola is, $\frac{x^2}{9}-\frac{y^2}{16}=1$
$\Rightarrow 16x^2-9y^2=144$
Hence, option 'A' is correct.