Showing that the locus of point $N$ is $x^2+y^2=a^2$

Question

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?

Answer 1

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$$ x^2 + y^2 = a^2. $$

Answer 2

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).

Answer 3

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 $\frac{1}{2}$, we have that

$$ON = \frac{1}{2}G F^\prime = \frac{1}{2}(GP + PF^\prime) = \frac{1}{2}(FP + PF^\prime) = \frac{1}{2}AB = OA$$