Proving a condition related to normal on ellipse

Question

Prove that the straight line $lx+my+n=0$ is a normal to the ellipse $x^2/a^2 +y^2/b^2=1$ if $a^2/l^2 +b^2/n^2 = (a^2 -b^2)^2/n^2$.

Answer 1

Using Article $267$ of Loney's The elements of coordinate geometry,

the equation of the normal of the given ellipse at $P(a\cos\phi, b\sin\phi)$ is $$x\cdot a\sec\phi-y\cdot b\csc\phi+b^2-a^2=0$$

So, $\displaystyle lx+my+n=0$ will be a normal if $$\frac l{a\sec\phi}=\frac m{-b\csc\phi}=\frac n{b^2-a^2}$$

$$\implies\cos\phi=\frac{na}{l(b^2-a^2)}\text{ and }\sin\phi=\frac{-nb}{m(b^2-a^2)}$$

Use $\displaystyle \cos^2\phi+\sin^2\phi=1$ to eliminate $\phi$

Answer 2

Hint: Let us consider a point $P=(x_0,y_0)$ on the upper half

$$y=|b|\sqrt{1-\frac{x^2}{a^2}}$$

of the given ellipse; in other words we consider a point $P=(x_0,|b|\sqrt{1-\frac{x_0^2}{a^2}})$. The lower half case is treated similarly. The tangent line through $(x_0,y_0)$ has analytic expression given by

$$y=y'(x_0)(x-x_0)+y(x_0),$$

while the normal is

$$y=-\frac{1}{y'(x_0)}(x-x_0)+y(x_0). $$

The line $lx+my+n=0\Leftrightarrow y=-\frac{l}{m}x-\frac{n}{m}$ (with $m,l\neq 0$) is then normal to the ellipse at $P$ if $$\frac{l}{m}=\frac{1}{y'(x_0)}=\frac{a^2\sqrt{1-\frac{x_0^2}{a^2}}}{|b|x_0},$$ $$\frac{n}{m}=-\frac{1}{y'(x_0)}x_0-y(x_0)=-\frac{a^2\sqrt{1-\frac{x_0^2}{a^2}}}{|b|}-|b|\sqrt{1-\frac{x_0^2}{a^2}}. $$

Using these two relations you should be able to prove the statement.

Answer 3

Besides to @Avitus solution, write down the leading vector of that normal line by evaluating:

$$\nabla F|_{(x_0,y_0)}=\frac{2y_0}{b^2}\mathbf{\hat{i}}+\frac{-2x_0}{a^2}\bf\hat{j}$$ and so the equation of that line is:

$$\text{L}:=\left(\frac{2y_0}{b^2}\right)x+\left(\frac{-2x_0}{a^2}\right)y+\left(\frac{2y_0x_0}{a^2}-\frac{2y_0x_0}{b^2}\right)$$