Question

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

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

Answer 1

Answer: (B)

Since the vertices are on the $y$ -axis (with origin at the mid point), the equation is of the form $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$.
As vertices are $(0, \pm 6), a=6$
$b^{2}=a^{2}\left(e^{2}-1\right)=36\left(\frac{25}{9}-1\right)=64$, so the required
equation of the hyperbola is:
$\Rightarrow \frac{y^{2}}{36}-\frac{x^{2}}{64}=1$