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