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Let's take from this question the equation for a rotated ellipse that is not centered at the origin:

$$\frac {((x-h)\cos(A)+(y-k)\sin(A))^2}{a^2}+\frac{((x-h) \sin(A)-(y-k) \cos(A))^2}{b^2}=1,$$

where $h, k$ and $a, b$ are the shifts and semi-axis in the $x$ and $y$ directions respectively and $A$ is the angle measured from $x$ axis.

I have also the general equation of a line:

$$ y = \alpha + \beta x $$

Assuming that $\alpha = 0$, how can I find the slope $\beta$ so that the line $y$ is a tangent of the ellipse? There should be two of them.

EDIT: I found also this question, but I don't understand the parameterization. It's basically the same question as mine, but I don't understand why the ellipse is defined by a system of equations and not a single equation; I also don't know what is $t$; it would be helpful if someone could just switch the parameterizations because the question seems to already have an answer.

J. Doe
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    The answer seems to be $$\beta=\frac{C\pm \sqrt{D}}{B},$$ where $$B=h^2+\left(\cos ^2A-1\right),b^2-\cos ^2A,a^2$$ $$C=h,k+\cos A,\sin A,b^2-\cos A,\sin A,a^2$$ $$D=\left(\left(1-\cos ^2A\right),b^2+\cos ^2A,a^2 \right),k^2+\left(2,\cos A,\sin A,b^2-2,\cos A,\sin A,a^2 \right),h,k+\left(\cos ^2A,b^2+\left(1-\cos ^2A\right),a^2 \right),h^2-a^2,b^2.$$ I found it by using dual conics. Do you want an answer detailing this approach? – Jan-Magnus Økland Apr 30 '23 at 16:00
  • @Jan-MagnusØkland Thanks, this seems excellent. Please post it as an answer. I think it is already detailed enough but of course it would maybe be helpful for future readers to learn about more details. – J. Doe Apr 30 '23 at 16:15

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The recipe is to take the dual conic (with matrix the adjugate matrix to the matrix of the conic) going through the point $(-\frac1{\alpha},\frac{\beta}{\alpha})$ corresponding to $y=\alpha+\beta x$ being tangent to the conic. Writing the resulting expression as a fraction with denominator having $\alpha^2$ as a factor, setting $\alpha=0$ in the numerator gives a quadratic in $\beta.$ I can flesh it out if you want. A CAS helps.

Jan-Magnus Økland
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