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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?

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    $\begingroup$ If $b=5$, then shouldn't it be $(y-5)^2\over 25$? Or perhaps it should be $(x-5)^2\over 2$? $\endgroup$
    – abiessu
    Commented May 16, 2014 at 17:28

1 Answer 1

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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 $b^2=c^2-a^2=...\:$.

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