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Find the equation of the ellipse that is concentric and tangent to the following hyperbolas:

$$\begin{align} -2x^2 + 9y^2 - 20x - 108y + 256 &= 0 \\ x^2 - 4y^2 + 10x + 48y - 219 &= 0 \end{align}$$

I did the math for both equations and the center is the same: $(-5,6)$. I have the equations of each: For the first: $$-2x^2 + 9y^2 - 20x - 108y + 256 = 0$$

$$-\frac{(x+5)^2}{9} + \frac{(y-6)^2}{2} = 1$$

For the 2nd: $$x^2 - 4y^2 + 10x + 48y - 219 = 0$$

$$\frac{(x+5)^2}{100} - \frac{(y-6)^2}{25} = 1$$

I know concentric means the new Ellipse has to have the same center, but I don't know how to make it tangent. Please help. Thanks

Blue
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user2092082
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1 Answers1

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Not a complete solution, but this approach will work:

An ellipse with centre at the point $(-5,6)$ would be

$$\frac{(x+5)^2}{a^2}+\frac{(y-6)^2}{b^2} = 1$$

Now change to a new set of axes ($u, v$) parallel to the $x,y$ axes, but with origin at the point $(-5,6)$. In other words, put $u = x+5$ and $v = y-6$. Referred to the new axes, the equations of the three conics become:

$$-\frac{u^2}{9} + \frac{v^2}{2} = 1,$$

$$\frac{u^2}{100} - \frac{v^2}{25} = 1,$$

$$\frac{u^2}{a^2}+\frac{v^2}{b^2} = 1$$

(Note that the values $a$ and $b$ are unchanged by the change in coordinates.)

Now, looking at a hyperbola $u^2/a^2-v^2/b^2 = \pm 1$ and an ellipse $u^2/c^2+v^2/d^2 = 1$, you can see (from a sketch) that the ellipse will be a tangent to the hyperbola only when they meet at a point on the $u$ or $v$ axis.

Can you finish it from there?

Old John
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