Question

Find the equation of hyperbola
whose center is $(-3,2)$, one vertex is $(-3,4)$ and eccentricity is $\frac{5}{2}$.

Answer 1

REF.Image.

a = distance b/w (3,2)& (-3,4)

$a=\sqrt{(4-2{)}^2}=2$

Now,

$b^2=a^2(e^2-1)=4(\frac{25}{4}-1)=21$

$e{q}^n$ of Hyperbola

$\frac{(x+3{)}^2}{4}-\frac{(y-2{)}^2}{21}=1$