The proposition is told to me without proof, may it is true but I cannot to believe it with my heart before I get the proof of it or the lights. I got some problems, however, after I try to proof it.
Let there is an ellipse $E(a,b)$ $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ then its auxiliary circle is $x^2 + y^2 = a^2$ and tangent of $(x_0,y_0)$ of the ellipse is $$\frac{x_0x}{a^2} + \frac{y_0y}{b^2} = 1.$$
Let $P$ is on both the auxiliary circle and the tangent, and $F = (-c,0)$ is the focus and $P_0 = (x_0,y_0)$, how to show $\overrightarrow{FP} \cdot \overrightarrow{P_0P} = 0$ ?
May one can show it by find out the point $P$. Let $m = \frac{x_0}{a^2}$ and $b = \frac{y_0}{b^2}$ thereby the equation become $mx + ny = 1$. Put it into the circle then $$\begin{aligned} (na)^2 &= (nx)^2 + (ny)^2 = (nx)^2 + (mx - 1)^2 \\ &= (n^2 + m^2)x^2 - 2mx + 1\end{aligned}$$
or $(n^2 + m^2)x^2 - 2mx + 1 - (na)^2 = 0$. Thereby the roots are $$x = \frac{2m \pm \sqrt{D}}{2(m^2 + n^2)} = \frac{m \pm \sqrt{n^2((m^2 + n^2)a^2 - 1)}}{m^2 + n^2}$$ where $$\begin{aligned} D &= 4m^2 - 4(m^2 + n^2)(1 - (na)^2) = 4(mna)^2 + 4n^2((na)^2 - 1)\\ &= 4n^2((m^2 + n^2)a^2 - 1) \end{aligned}$$
If bring the $\frac{x_0}{a^2}$ and $\frac{y_0}{b^2}$ back into the equation of the roots, how can one easily look out $\overrightarrow{FP} \cdot \overrightarrow{P_0P} = 0$ ?
Or, to find out the solutions of equation $\overrightarrow{FP} \cdot \overrightarrow{P_0P} = 0$ first, $(x - x_0, y - y_0) \cdot (x + c,y) = 0$ it is. Thereby $(x + c)(x - x_0) + y(y - y_0) = x^2 + y^2 + (c - x_0)x - y_0y - cx_0 = 0$. I would like to see $a^2 = (x_0 - x)x + y_0y + cx_0 $ but the $x$ and $y$ are variables such that look like impossible.
Let $E(a,b)$ is set of points of the ellipse such that $p \in E(a,b)$ have form $(a\cos \theta, b\sin \theta)$. And let there is a matrix such that $C(r) := E(r,r) \cong E(a,b)$ which $$A = \begin{bmatrix} \alpha & 0 \\ 0 & \beta \end{bmatrix}.$$ Thereby $$\alpha = \frac{a}{r},\ \beta = \frac{b}{r} \ \mathrm{and}\ A^{-1} = \begin{bmatrix} \alpha^{-1} & 0 \\ 0 & \beta^{-1} \end{bmatrix}. $$
Therefore $Ap' = (-a\dot{\theta}\sin \theta, b\dot{\theta}\cos \theta) = (Ap)'$ where $p \in C(r)$. Can it be used to bring the question into problem of circle and its tangent, then carry the solution back to the question of ellipse and its tangent ?
