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.
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.