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Lets say I have a circle with radius "r" and center "x1,y1" in the X-Y plane. If I rotate the circle by "theta" degrees about the Y axis, I would see an ellipse from the X-Y plane.

How could I obtain the equation for this ellipse?

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Plane rotated $\theta$ around Y axis: $$ \mathbb{P} : X x_1 \sin \theta + Z x_1 \cos \theta = 0 $$ Center and radius must hold: $$ \xi : (x + x_1 \cos \theta ) ^2 + (y - y_1 ) ^2 + (z - x_1 \sin \theta ) ^2 = r^2 $$ Intersecting both surfaces and Parametrizing ($\phi$) the curve: $$ X = (r \sin \phi - x_1 ) \cos \theta \rightarrow Z = - (r\sin\phi - x_1)\sin\theta \quad Y = r\cos \phi + y_1 $$

$$ \frac{(X + x_1 \cos \theta)}{\cos \theta} = r \sin\phi \\ (Y - y_1) = r \cos\phi $$ $$ \zeta : \frac{(X + x_1 \cos \theta)^2}{r^2 \cos^2\theta} + \frac{(Y-y_1)^2}{r^2} = 1 $$

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The original circle has equation $$(x-x_1)^2+(y-y_1)^2=r^2$$

The ellipse with the same centre and with semiaxes $a$ and $b$ has equation $$\frac{(x-x_1)^2}{a^2}+\frac{(y-y_1)^2}{b^2}=1$$

When you rotate the circle by angle $\theta$ about the $y$ axis, the coordinates of the centre become $$(x_1\cos \theta, y_1)$$ and the semiaxes become $r\cos \theta$ and $r$.

So you can write out the equation of the ellipse. I hope that helps.

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