I'm trying to generalize this proof ($\alpha=90^\circ$ then $C_2$ is a point) to prove the following
- A chord $BC$ of a conic $C_1$ which subtends a constant angle $\angle BAC=\alpha$ at a given point $A$ on $C_1$ envelopes a conic $C_2$.
- Two conics $C_1,C_2$ have double contact (This answer mentions the definition of double contact).
- Let $T_1,T_2$ be tangency points of $C_1,C_2$. The slope of $AT_1,AT_2$ are $\pm i$.
- Let $D$ be the tangency point of $BC$ with $C_2$, let $E$ be the pole of the line $BC$ wrt $C_1$, then are the lines $AD$ and $AE$ always symmetric wrt bisector of $\angle BAC$
Some thoughts:
- This answer mentioned that two conics having double contact can be mapped to two concentric circles after a complex projective transformation. So it suffices to show that $C_1,C_2$ can be mapped to two concentric circles after a complex projective transformation.
- Let $B$ be a point on $C_1$, it is easy to show that the map $B\mapsto C$ such that $\angle BAC=\alpha$ a projective transformation on $C_1$: Let four points $B_1,B_2,B_3,B_4$ be on $C_1$, then the cross ratio of the lines $AB_1,AB_2$ and $AB_3,AB_4$ is preserved under the map $B\mapsto C$ such that $\angle BAC=\alpha$. So the map $B\mapsto C$ is a projective transformation on $C_1$ with fixpoints $T_1,T_2$.
- Point $A(x_0, y_0)$ is a fixed point on the conic $C_1:ax^2 + by^2 = 1$ ($ab \neq 0$). Points $B$ and $C$ are moving on $C_1$, such that $\tan\angle BAC = t$. Then the envelope of the line $BC$ is the conic $C_2:$
$$4ab(t^2+1)(ax^2+by^2-1) + t^2((a-b)(a x x_0 - b y y_0) + a + b)^2 = 0.$$
so the line $T_1T_2$ is $(a-b)(a x x_0 - b y y_0) + a + b=0$. Interestingly it doesn't depend on $t$.
$C_1$ can be put into the above form $ax^2+by^2=1$ except when it is a parabola. In this post I calculated the equation of $C_2$ when $C_1$ is a parabola.
