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Question: A point $P(acostheta,bsintheta)$ lies on an ellipse with equation $$varepsilon:fracx^2a^2+fracy^2b^2=1.$$The tangent to the ellipse $varepsilon$ at point $P$ is perpendicular to a straight line $l$ which has passed through its focus and intersected at point $N$. Show that the equation of locus of point $N$ is $x^2+y^2=a^2$.

Suppose the coordinate point of the focus is $F(c,0)$. To find the point $N$ parametrically, I have to deal with the following simultaneous equations.
$$begincases
y-b sin theta=-dfracba cot theta , (x- a cos theta)\y=dfracbatan theta ,(x-c)
endcases$$
where the first equation is the equation of tangent line at point $P$, and the second equation represents the perpendicular line $l$ that passes through the focus of the ellipse $varepsilon$ and point $N$.

Solving the above equation for $x$ and $y$ in terms of $a$, $b$ and $c$ is quite complicated and outsmarted because it also involves a new variable $theta$. Any pretty way to deal with it?

Denote $P$ by $(x_P, y_P)$ instead, then $dfracx_P^2a^2 + dfracy_P^2b^2 = 1$ and the tangent line at $P$ is$$
t_P: fracx_P xa^2 + fracy_P yb^2 = 1. tag1
$$
Because $l$ is perpendicular to $t_P$, then $l$ is given by$$
l: -fracy_P xb^2 + fracx_P ya^2 = C_P,
$$
where $C_P$ is a constant depending on $P$. Given that $l$ passes through $(c, 0)$, thus $C_P = -dfracc y_P b^2$ and$$
l: -fracy_P xb^2 + fracx_P ya^2 = -fracc y_Pb^2. tag2
$$

Now, $(1)^2 + (2)^2$ yields$$
1 + fracc^2 y_P^2b^4 = left( -fracy_P xb^2 + fracx_P ya^2 right)^2 + left( -fracy_P xb^2 + fracx_P ya^2 right)^2 = left( fracx_P^2a^4 + fracy_P^2b^4 right) (x^2 + y^2). tag3
$$
Since $x_P^2 = a^2 left( 1 - dfracy_P^2b^2 right)$ and $a^2 - b^2 = c^2$, then$$
fracx_P^2a^4 + fracy_P^2b^4 = frac1a^2 left( 1 - fracy_P^2b^2 right) + fracy_P^2b^4 = frac1a^2 left( 1 - fracy_P^2b^2 + fraca^2 y_P^2b^4 right) = frac1a^2 left( 1 + fracc^2 y_P^2b^4 right)
$$
and (3) becomes$$
x^2 + y^2 = a^2.
$$

Preliminaries:-

(1) For the standard ellipse ($epsilon : dfrac x^2a^2 + dfrac y^2b^2 = 1$), its foci are $(ae, 0)$ and $(–ae, 0)$; where $e$ is the eccentricity and is related to $a$ and $b$ by $b^2 = a^2(1 – e^2)$.

(2) If $L$ is tangent to $epsilon$ at $P[theta]$, then the equation of $L$ is $dfrac x cos thetaa + dfrac x sin thetab = 1$.

(3) An alternate form of $L$ is $y = mx + c$ for some $m$ and $c$.

(4) Eliminating $c$ from the equations found in (2) and (3), we have $c = pm sqrt m^2a^2 + b^2$. That is, the equation of $L$ is $L : y – mx = pm sqrt m^2a^2 + b^2$.

The main part:-

(5) If $N$ is the line that passes thro’ (ae, 0) and perpendicular to $L$, then ….. $N : my + x = ae$.

(6) Adding the Squares of both sides of (4) and (5), we get $(1 + m^2)(x ^2 + y^2) = b^2 + a^2e^2 + m^2a^2$.

After replacing the $b^2$ in (6) by the relation stated in (1), the required result follows when we eliminate $(1 + m^2)$ from both sides of (6).

This is the auxiliary circle property of the ellipse, and I think the classical approach is beautiful, plus it avoids all of the equation crunching.

Suppose the ellipse has foci $F$ and $F^prime$ and center $O$. We need two famous properties of ellipses:

1. Locus definition: $FP + PF^prime = AB$
   


2. Optical property: $angle FPN cong angle F^prime P M$

Then, $GP = FP$ since $FNP cong GNP$. Also, since $FNO sim F G F^prime$ with similarity ratio $frac12$, we have that

$$ON = frac12G F^prime = frac12(GP + PF^prime) = frac12(FP + PF^prime) = frac12AB = OA$$