I've tried to solve this question, but I'm stuck. The question is:
Find the equation in standard form for the hyperbola centered at $(5,5)$ with one focus at $(-7,5)$ and eccentricity at $\frac{3}{2}$.
When I tried to solve this, I got $a$ to be $2$, $c$ to be $3$, and $b$ to be $5$.
The equation I for was:
$$\frac{(x-5)^2}{4}-\frac{(y-5)^2}{5}=1.$$
However, the numbers don't seem correct.
Any ideas on where I went wrong?
Hint: The distance from the center $(5,5)$ to the focus $(-7,5)$ is $c$ and equals $12$, Thus $\varepsilon=\frac{3}{2}=\frac{12}{a}$ that implies $a=...\:$. Then $b^2=c^2-a^2=...\:$.