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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 ?

Mark
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Dinesh
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  • There's an easier way you do not need the complex number, express it as quadratic having four roots, use vietas . you'd see it. Anyway ellipse in complex plane is given by $|z-z_1|+|z-z_2|=2a$ where $z_1$ and $z_2$ are foci. – Someone May 14 '15 at 06:17
  • @Rory I corrected it – Dinesh May 14 '15 at 06:20
  • @Mann I got the ans but I'd thought that via complex it might be shorter. And I tried to make a fourth degree equation in $z$ and thru it I solved first bracket easily, but I can't go for second one. You see after simplifying it becomes $\frac{m^2}{m^2+1}$ where m is a root and so we cannot transform the equation easily into the equation having these as roots – Dinesh May 14 '15 at 06:24
  • http://chat.stackexchange.com/rooms/23542/discussion-between-mann-and-adg , join this. ^^ @Dinesh , won't be to good to carry on in comment section. – Someone May 14 '15 at 06:25

1 Answers1

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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.

Ng Chung Tak
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