<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] Denote $P$ by $(x_P, y_P)$ instead, then $\dfrac{x_P^2}{a^2} + \dfrac{y_P^2}{b^2} = 1$ and the tangent line at $P$ is$$
t_P: \frac{x_P x}{a^2} + \frac{y_P y}{b^2} = 1. \tag{1}
$$
Because $l$ is perpendicular to $t_P$, then $l$ is given by$$
l: -\frac{y_P x}{b^2} + \frac{x_P y}{a^2} = C_P,
$$
where $C_P$ is a constant depending on $P$. Given that $l$ passes through $(c, 0)$, thus $C_P = -\dfrac{c y_P }{b^2}$ and$$
l: -\frac{y_P x}{b^2} + \frac{x_P y}{a^2} = -\frac{c y_P}{b^2}. \tag{2}
$$
Now, $(1)^2 + (2)^2$ yields$$
1 + \frac{c^2 y_P^2}{b^4} = \left( -\frac{y_P x}{b^2} + \frac{x_P y}{a^2} \right)^2 + \left( -\frac{y_P x}{b^2} + \frac{x_P y}{a^2} \right)^2 = \left( \frac{x_P^2}{a^4} + \frac{y_P^2}{b^4} \right) (x^2 + y^2). \tag{3}
$$
Since $x_P^2 = a^2 \left( 1 - \dfrac{y_P^2}{b^2} \right)$ and $a^2 - b^2 = c^2$, then$$
\frac{x_P^2}{a^4} + \frac{y_P^2}{b^4} = \frac{1}{a^2} \left( 1 - \frac{y_P^2}{b^2} \right) + \frac{y_P^2}{b^4} = \frac{1}{a^2} \left( 1 - \frac{y_P^2}{b^2} + \frac{a^2 y_P^2}{b^4} \right) = \frac{1}{a^2} \left( 1 + \frac{c^2 y_P^2}{b^4} \right)
$$
and (3) becomes$$
x^2 + y^2 = a^2.
$$
</s>