<s>[INST] Showing that the locus of point $N$ is $x^2+y^2=a^2$ 
Question: A point $P(a\cos\theta,b\sin\theta)$ lies on an ellipse with equation $$\varepsilon:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$$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.
$$\begin{cases}
y-b \sin \theta=-\dfrac{b}{a} \cot \theta \, (x- a \cos \theta)\\y=\dfrac{b}{a}\tan \theta \,(x-c)
\end{cases}$$
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?
 [/INST] Preliminaries:-
(1) For the standard ellipse ($\epsilon : \dfrac {x^2}{a^2} + \dfrac {y^2}{b^2} = 1$), 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 $\dfrac {x \cos \theta}{a} + \dfrac { x \sin \theta}{b} = 1$.
(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). 
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