If $4ax+3by+12c$ is normal to ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, then how are $c$, $a$, $e$ related?

Question

Here we have equation of normal of ellipse, $4ax+3by=12c$ and ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

What is the relation between $(c, a, e)$?

(a) $c=a^2e^2\quad$ (b) $c=5a^2e^2\quad$ (c) $5c=a^2e^2\quad$ (d) None of these

The answer is (c) $5c=a^2e^2$.

I tried to solved it with formula of normal and by putting normal points in it. ($ae,\pm\dfrac{b^2}{e}$), so here $\dfrac{a^2x}{x_1}+\dfrac{b^2y}{y_1}=a^2-b^2$, but couldn't find the relation.

Answer 1

Here's my try

The condition for a line $Ax+By+C=0$ to be normal to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is

$$\frac{a^2}{A^2}+\frac{b^2}{B^2}=\frac{(a^2-b^2)^2}{C^2}$$

Now, your line is $4ax+3by-12c=0$ and ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, on comparing we get,

1. $A=4a$

2. $B=3b$

3. $C=-12c$

Also using the fact that $b^2=a^2(1-e^2)$ to get $a^2-b^2=a^2e^2$;

$$\frac{a^2}{16a^2}+\frac{b^2}{9b^2}=\frac{(a^2e^2)^2}{144c^2}$$

$$5c=a^2e^2$$

So option (c) is correct.

Extra- the parametric coordinates of an ellipse of the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ are $(acos\theta,bsin\theta)$ and the equation of normal to the ellipse in parametric form is $\frac{ax}{cos\theta}+\frac{by}{sin\theta}=a^2-b^2$. Compare this normal equation with Ax+By+C=0, we get $sin\theta=\frac{bC}{B(a^2-b^2)}$ and $cos\theta=\frac{aC}{A(a^2-b^2)}$. Now use the usual identity $sin^2\theta+cos^2\theta=1$ to get to the condition mentioned above.