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?

• Can we avoid solving simultaneous equations above? – weilam06 Mar 29 '19 at 14:41

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

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

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

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).