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Give the equation of the ellipse $x^2+2y^2-6x+16y+9=0$ after reflection in the line $y=-x$.

I completed the square and obtained $$\frac{(x-3)^2}{32}+\frac{(y+4)^2}{16}=1$$

Now I changed $y$ and $x$ and then replaced $x$ with $-x$ to obtain $$\frac{(y-3)^2}{32}+\frac{(-x+4)^2}{16}=1$$

My teacher says I am not correct and that I should replace $x$ with $-y$ and $y$ with $-x$ (why is this) whereas my answer book says the answer should be $\frac{(x-4)^2}{16}+\frac{(y+5)^2}{32}$.

It's driving me crazy, all those $x$ and $y$, could someone clarify for me?

Dr. Heinz Doofenshmirtz
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1 Answers1

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Plot any point $(x_1,y_1)$ in the plane. Then reflect that point across the line $y=-x$ and see where the reflection of the point is. The reflection will be at $(x_1',y_1') = (-y_1,-x_1)$.

Note that not the signs of both coordinates are changed, not just the $x$ coordinate.

The result (with the correct signs of both coordinates) is that the reflected ellipse has formula

$$\frac{(-y-3)^2}{32}+\frac{(-x+4)^2}{16}=1$$

Now observe that $(-x+4)^2 = (x-4)^2$ and $(-y-3)^2 = (y+3)^2.$ I don't know why your book says $y+5$ instead of $y+3$; you completed the square correctly and the $x-3$ in your first (unreflected) formula is correct.

David K
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