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Normals at $4$ points on an ellipse centred at the origin are drawn. They intersect at $(a, b)$. We need to find the mean position of the points.

Well I've the standard equations of the normals to an ellipse, I'm not being able to carry the relation. I tried finding relations between $a$, $b$ and the ellipse parameters, hoping to make sone development, in vain.

How should I solve the problem?

My attempts:

Equation of normal: $$\frac{(b^2) y}{y_1}-\frac{(a^2) x}{x_1}=-a^2+b^2$$

where $b$ is the semi minor axis and $a$ is the semi major axis. $x_1$ and $y_1$ are the coordinates at which the normal is drawn on the ellipse.

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  • $\begingroup$ How about the four points where the ellipse intersects the coordinate axes and whose normals are the coordinate axes and these intersect at the origin? $\endgroup$ – Somos Dec 9 '17 at 5:07
  • $\begingroup$ By "mean position", you mean their barycenter ? Besides, you can transfer your problem onto the "evolute curve" of the ellipse which is an astroid ; see the answer by Achille Hui in (math.stackexchange.com/q/609351). $\endgroup$ – Jean Marie Dec 9 '17 at 7:01
  • $\begingroup$ Your question is ill-posed : instead of saying "Normals at 4 points on an ellipse centered at the origin are drawn. They intersect at (a, b). " you should say "from a point $(a,b)$ (or $(x_1,y_1)$ you must choose), four normals to the ellipse are drawn" (is it always possible to have 4 of them, that is the question) $\endgroup$ – Jean Marie Dec 9 '17 at 9:26
  • $\begingroup$ See askiitians.com/iit-jee-coordinate-geometry/… $\endgroup$ – lab bhattacharjee Dec 9 '17 at 10:40
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The equation of the normal passing through a point $(x_0,y_0)$ of the ellipse and a generic point $(x,y)$ is $$ \tag{1} x_0b^2(y-y_0)=y_0a^2(x-x_0). $$ This must be combined with the ellipse equation: $$ \tag{2} b^2x_0^2+a^2y_0^2=a^2b^2, $$ where $a$ and $b$ are the semi-axes. We can solve $(1)$ for $y_0$ and plug the result into $(2)$, thus obtaining a resolvent quartic equation in $x_0$: $$ \tag{3} (a^2b^2 -b^2x_0^2)[a^2x-(a^2-b^2)x_0]^2-a^2b^4x_0^2y^2=0. $$ We know by hypothesis that this equation admits $4$ real solutions for $x_0$. The sum $S_x$ of the solutions is given by $-A_3/A_4$, where $A_3$ is the coefficient of $x_0^3$ in the polynomial on the left hand side and $A_4$ is the coefficient of $x_0^4$. By expanding $(3)$ we can find this to be: $S_x=2a^2x/(a^2-b^2)$.

We can repeat the same process, solving $(1)$ for $x_0$ and plugging the result into $(2)$, to obtain a resolvent quartic equation in $y_0$ with four real solutions. The sum $S_y$ of the solutions turns out to be: $S_y=-2b^2y/(a^2-b^2)$.

Finally, the coordinates of the centroid $G$ of the four points are given by $(S_x/4,S_y/4)$, that is: $$ G={1\over2(a^2-b^2)}\big(-a^2x,\,b^2y\big). $$ Notice that $P=(x,y)$ in the above formula is the point where the four normals intersect.

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