Geometrical place of points on ellipses closest to a point

Question

We have fixed point $A(x_0,y_0)$ and family of ellipses with fixed focuses $(c,0)$ and $(-c,0)$. For each of the ellipses we can find point $E(x_1,y_1)$ on the ellipse closest to $A$.

I'm interested to find geometrical place of $E$s, actually the equation including $x_1,y_1,x_0,y_0,c$

Obviously $A$ belongs to the curve, so as point $(x_0,0)$ if $|x_0|<c$. If $x_0>c$ then $(c,0)$ is on the curve

Any idea how to set up the equation?

Answer 1

I was trying myself to answer, I supposed that the tangent in $E$ on the ellipse is also the tangent in $E$ on a circle with center $(x_0,y_0)$.

If ellipse equation is $\frac{x^2}{b^2+c^2}+\frac{y^2}{b^2}=1$ and the circles equation is $(x-x_0)^2+(y-y_0)^2=r^2$ then equations $\frac{xx_1}{b^2+c^2}+\frac{yy_1}{b^2}=1$ and $(x-x_0)(x_1-x_0)+(y-y_0)(y_1-y_0)=r^2$ are both equations of the same line

So coefficients near x and y are proportional $$\frac{x_1}{b^2+c^2}:(x_1-x_0)=\frac{y_1}{b^2}:(y_1-y_0)\\ x_1(y_1-y_0)b^2=y_1(x_1-x_0)(b^2+c^2)\\ (x_0y_1-x_1y_0)b^2=y_1(x_1-x_0)c^2\\ b^2=\frac{y_1(x_1-x_0)c^2}{x_0y_1-x_1y_0}$$ Now substitute $b^2$ into ellipses equation $$\frac{x_1^2}{b^2+c^2}+\frac{y_1^2}{b^2}=1\ \ /*\frac{c^2}{x_0y_1-x_1y_0}\\ \frac{x_1^2}{y_1(x_1-x_0)+x_0y_1-x_1y_0}+\frac{y_1^2}{y_1(x_1-x_0)}=\frac{c^2}{x_0y_1-x_1y_0}\\ \frac{x_1}{y_1-y_0}+\frac{y_1}{x_1-x_0}=\frac{c^2}{x_0y_1-x_1y_0}\\ x_1(x_1-x_0)(x_0y_1-x_1y_0)+y_1(y_1-y_0)(x_0y_1-x_1y_0)=c^2(x_1-x_0)(y_1-y_0)$$

Answer 2

A parametric equation for this family of ellipses is

$$\frac{x^2}{t^2} + \frac{y^2}{t^2 - 4c^2} = \frac{1}{4}$$

and the lagrangian is

$$(x-x_0)^2 + (y-y_0)^2 - \lambda (\frac{x^2}{t^2} + \frac{y^2}{t^2 - 4c^2} - \frac{1}{4})$$

Solving for stationary points gives $x = \frac{t^2}{t^2 - \lambda}x_0$, $y = \frac{t^2 - 4c^2}{t^2 - 4c^2 - \lambda}y_0$, and the only thing left is to solve

$$\frac{t^2}{(t^2 - \lambda)^2}x_0^2 + \frac{t^2 - 4c^2}{(t^2 - 4c^2 - \lambda)^2}y_0^2 = \frac{1}{4} $$

for $\lambda$. Quite an unpleasant quartic.

Answer 3

It is not so complicated.

Equation of normal to an ellipse $(x/a)^2 + (y/b)^2 =1$ in straight line form $ y = m x +c $ is

$$ y = mx \pm m (a^2-b^2) /(a^2+b^2 m^2) $$

or

$$ y = mx \pm c^2/(a^2(1+m^2)-c^2 m^2) $$

or when passing through $ (x_0,y_0)$ becomes

$$ y_0 = m x_0 \pm c^2/(a^2(1+m^2)-c^2 m^2) $$

which is asimple equation helping to find $a$ for a given $m$ or vice-versa.