Q:

Tangent to hyperbola meeting on ellipse

Can anyone prove that the two tangent lines to hyperbola that meet on a confocal ellipse make the same angle with the tangent of the ellipse at the point of intersection?
Thanks

A:

Alternative formulation. Let us have an ellipse and a hyperbola that share the same pair of foci $F_1$ and $F_2$. Let point $E$ be a point on the ellipse and let the lines $t_1$ and $t_2$ be the two tangents to the hyperbola that pass through the point $E$. Let $H_1$ and $H_2$ be the points where the lines $t_1$ and $t_2$ touch the hyperbola respectively. 
Let $t_E$ be the line tangent to the ellipse at point $E$ with $E1$ and $E2$ two arbitrary points chosen on $t_E$ so that $E$ is between $E_1$ and $E_2$, as shown on the figure. Prove that 
$$\angle \, E_1EH_1 = \angle \, E_2EH_2$$
$$ $$
 
Proof. Let $F_1$ and $F_2$ be the foci of the hyperbola in question. Let $H_1$ and $H_2$ be two points from the hyperbola and let lines $t_1$ and $t_2$ be the two tangents to the hyperbola at the points $H_1$ and $H_2$ respectively. Take the unique ellipse with foci $F_1$ and $F_2$ that passes through point $E = t_1 \cap t_2$. Indeed, this is the ellipse, formed by all points $X$ in the planes such that $$F_1X + XF_2 = F_1E + EF_2$$ The line $t_E$ is the tangent line to the ellipse at the point $E$ and points $E_1$ and $E_2$ are two arbitrary points chosen on $t_E$ so that $E$ is between $E_1$ and $E_2$, as shown on the figure. 

By the reflective properties of ellipses, 
$$\angle \, E_1EF_1 = \angle \, E_2EF_2 = \alpha$$
By the reflective properties of hyperbolas, 
$$\angle \, F_1H_1E = \angle \, F_2H_1E \,\, \text{ and } \,\, 
\angle \, F_1H_2E = \angle \, F_2H_2E$$ 
Let the line through point $F_1$ orthogonal to $t_1 = H_1E$ intersects $H_1F_2$ at point $S_1$, and let he line through point $F_2$ orthogonal to $t_2 = H_2E$ intersects $H_2F_1$ at point $S_2$. Since $t_1$ is the angle bisector of $\angle \, F_1H_1F_2$ and $t_2$ is the angle bisector of $\angle\, F_1H_2F_2$, the two triangles $F_1H_1S_1$ and $F_2H_2S_2$ are isosceles with $H_1F_1 = H_1S_1$ and $H_2F_2 = H_2S_2$ respectively. 
By the properties of the hyperbola, $$F_1S_2 = F_1H_2 - H_2S_2 = F_1H_2 - H_2F_2 = F_2H_1 - H_1F_1 = F_2H_1 - H_1S_1 = F_2S_1$$
Furthermore, by construction, line $t_1 = H_1E$ is the orthogonal bisector of segment $F_1S_1$, and  line $t2 = H_2E$ is the orthogonal bisector of segment $F_2S_2$. Therefore, $EF_1 = ES_1$ and $EF_2 = ES_2$. Combined with the fact that $F_1S_2 = F_2S_1$, all of these identities yield that triangles $EF_1S_2$ and $EF_2S_1$ are congruent. Hence, $$\angle \, F_1ES_2 = \angle \, S_1EF_2 = \beta$$ Let $\angle \, S_2ES_1 = \theta$. Then $$\angle \, F_1ES_1 = \angle \, F_1ES_2 + \angle \, S_2ES_1 = \beta + \theta = \theta + \beta = \angle \, S_2ES_1 + \angle \, S_1EF_2 = \angle \, S_2EF_2$$ and for that reason $$\angle \, F_1EH_1 = \frac{1}{2} \, \angle \, F_1ES_1 = \frac{1}{2} \, (\beta + \theta) =  \frac{1}{2} \, \angle \, S_2ES_2 = \angle \, F_2EH_2$$
Finally, 
$$\angle \, E_1EH_1 = \angle \, E_1EF_1 + \angle \, F_1EH_1 =\alpha + \frac{1}{2}(\beta + \theta)$$ and
$$\angle \, E_2EH_2 = \angle \, E_2EF_2 + \angle \, F_2EH_2 =\alpha + \frac{1}{2}(\beta + \theta)$$ Hence
$$\angle \, E_1EH_1 = \angle \, E_2EH_2$$

