Question

The equation of hyperbola with vertices at $(0,\pm 6)$ and $e=\frac{5}{3}$
(where $e$ is eccentricity)

• A. $\frac{y^2}{36}-\frac{x^2}{64}=1$
• B. $\frac{x^2}{64}-\frac{y^2}{36}=1$
• C. $\frac{x^2}{36}-\frac{y^2}{64}=1$
• D. $\frac{y^2}{35}-\frac{x^2}{25}=1$

Answer 1

The correct option is A

$\frac{y^2}{36}-\frac{x^2}{64}=1$

Since vertices are on $y-$axis (with origin as the centre)
therefore equation of hyperbola will be
$\frac{y^2}{b^2}-\frac{x^2}{a^2}=1$

Distance between centre and vertices $=b=6$
$\Rightarrow e^2=1+\frac{a^2}{b^2}$
$\Rightarrow \frac{25}{9}-1=\frac{a^2}{b^2}$
$\Rightarrow a^2=\frac{16\left(36\right)}{9}=64$
$\Rightarrow \frac{y^2}{36}-\frac{x^2}{64}=1$