complex nos in ellipse. I was practising some ques on ellipses when I came a criss this question:
If normal at four points $(x_1,y_1)$..... on the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$  are concurrent then find the value of  $$(x_1+x_2+x_3+x_4)\left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}\right)$$
I know how to solve this question by co ordinate geometry formulas, but I want to do it with complex Nos. I let $z=\cos(\theta)$ and replaced $\sin$ and $\cos$ in the equation of normal by $z$ but I cannot simplify the second bracket. Can anybody help me to show how its done ?
 A: Equation of tangent at $(x_k,y_k)$,
$$\frac{x_kx}{a^2}+\frac{y_ky}{b^2}=1$$
Equation of normal at $(x_k,y_k)$,
$$\frac{y_k}{b^2}(x-x_k)-\frac{x_k}{a^2}(y-y_k)=0$$
If $(X,Y)$ is the common point of the four normals, then
$$\frac{y_k}{b^2}(X-x_k)-\frac{x_k}{a^2}(Y-y_k)=0$$
Hence, all the points $(x_k,y_k)$ lie on another conic:
$$\frac{y}{b^2}(X-x)-\frac{x}{a^2}(Y-y)=0$$
Re-arrange, $$y=\frac{b^2 Y x}{a^2(x-X)+b^2x}$$
Put back into the ellipse:
\begin{align}
  0 &=(a^2+b^2)^2 x^4-2a^2X(a^2+b^2)x^3-a^2[(a^2+b^2)^2-a^2 X^2+b^2 Y^2]x^2 \\
  & \qquad +2 a^4X(a^2+b^2)x-a^6 X^2
\end{align}
By Vieta's formulae,
\begin{align}
  (x_1+x_2+x_3+x_4)
  \left(
    \frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}
  \right)
  &= \frac{\displaystyle
           \left( \sum x_i \right)
           \left(\sum x_i x_j x_k \right)}{\displaystyle \prod x_i} \\[5pt]
  &=
  \frac{
    \left( \dfrac{2a^2X}{a^2+b^2} \right)
    \left( -\dfrac{2a^4X}{a^2+b^2} \right)}
       {-\dfrac{a^6 X^2}{(a^2+b^2)^2}} \\[5pt]
  &= 4
\end{align}
providing $X\ne 0$.
It won't be the case for oblique central conics.  See another answer here.
