Could someone tell me where I can find the generic equation of the $\alpha$-isoptic of a conic section of this form: $$\Gamma: ax^2+2bxy+cy^2+2dx+2ey+f=0$$ I searched in many links and PDFs but I found nothing.
($\alpha$-isoptic curve is the locus of the points through which the tangents to the conic pass and which intersect forming an angle $\alpha$)
I solved it by myself. Omitting the steps the formula is this:
Let be $I_1,I_2$ and $I_3$ the invariants of the conic section $$\Gamma(x,y):ax^2+2bxy+cy^2+2dx+2ey+f=0$$ If $$\mathbf{A}=\begin{pmatrix}a&b&d\\b&c&e\\d&e&f\end{pmatrix}\qquad\mathbf{B}=\begin{pmatrix}a&b\\b&c\end{pmatrix}$$ We define $$I_3=\det(\mathbf{A})\qquad I_2=\det(\mathbf{B})\qquad I_1=\text{tr}(\mathbf{B})$$ And $$(x_0,y_0)=\left(\dfrac{cd-eb}{b^2-ac}, \dfrac{ae-db}{b^2-ac}\right)$$ is the center of $\Gamma(x,y)$.
The $\alpha$-isoptic curve of $\Gamma(x,y)=ax^2+2bxy+cy^2+2dx+2ey+f$ is: $$\left[\tan(\alpha)\frac{ I_2}{2}\left((x-x_0)^2+(y-y_0)^2+\frac{I_1 I_3}{I_2^2}\right)\right]^2=-I_3\cdot(ax^2+2bxy+cy^2+2dx+2ey+f)$$