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Problem:

Find the equation for the hyperbola which has foci $$F_1 = (-1, 3)$$ $$F_2 = (3,3) $$ and eccentricity $$\varepsilon = 2$$

Hint: Use a translation which moves the foci to the x-axis.

My attempt:

Using a simple translation $$\textbf{R} = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & -3 \\ 0 & 0 & 1\end{bmatrix}$$ I have translated the hyperbola 3 units down, such that the foci are on the x-axis.

I am not able to progress from here, and I can't find any formulae to help me.

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As $y=3$ contains the foci, it also contains the major axis

If the equation is $\dfrac{(x-\alpha)^2}{a^2}-\dfrac{(y-\beta)^2}{b^2}=1$

As the center is the midpoint of the foci, we have $2\alpha=-1+3,2\beta=3+3$

Now the coordinates of the foci are $(\alpha\pm a\varepsilon,\beta)$

So, $1+2a=3,1-2a=-1\implies a=1$

We know $b^2=a^2(\varepsilon^2-1)$

Hope you can take it form here

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  • $\begingroup$ Just to be sure, in your first line, the left-hand side is supposed to be a subtraction, and not an addition, right? $\endgroup$ – Alec Jan 18 '15 at 17:00
  • $\begingroup$ @Aleksander, Rectified. Thanks $\endgroup$ – lab bhattacharjee Jan 18 '15 at 17:37
  • $\begingroup$ Thanks! I'm able to follow this and solve the problem, but I was wondering; how do you find that $$b^2 = a^2(\varepsilon^2-1)$$? $\endgroup$ – Alec Jan 18 '15 at 17:53
  • $\begingroup$ Never mind. Slow day. I guess it's from $$\varepsilon = \frac{\sqrt{a^2-b^2}{}}{a}$$ $\endgroup$ – Alec Jan 18 '15 at 18:04
  • $\begingroup$ @Aleksander, See en.wikipedia.org/wiki/Eccentricity_(mathematics)#Values $\endgroup$ – lab bhattacharjee Jan 19 '15 at 4:42

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