Question

If the normal at any point $P$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ meets the coordinate axes in $G$ and $g$ respectively, Then find the ratio $\frac{PG}{Pg}$

• A. $\frac{a^2}{b^2}$
• B. $\frac{a^4}{b^4}$
• C. $\frac{b^4}{a^4}$
• D. $\frac{b^2}{a^2}$

Answer 1

The correct option is D $\frac{b^2}{a^2}$

Let $P(a\cos \theta ,b\sin \theta )$ be a point on ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$


Equation of normal at point $P:$
$\frac{ax}{\cos \theta }-\frac{by}{\sin \theta }=a^2-b^2$
$\Rightarrow ax\sec \theta -by\csc \theta =a^2-b^2$

For coordinates of $G,$ Put $y=0.$
$ax\sec \theta =a^2-b^2$
$x=\frac{a^2-b^2}{a}\cos \theta$
So, coordinates of $G$ are $(\frac{a^2-b^2}{a}\cos \theta ,0)$

For coordinates of $g,$ Put $x=0.$
$-by\csc \theta =a^2-b^2$
$y=-\frac{a^2-b^2}{b}\sin \theta$
So, coordinates of $G$ are $(0,-\frac{a^2-b^2}{b}\sin \theta )$

Now, $P{G}^2=(a\cos \theta -\left(\frac{a^2-b^2}{a}\right)\cos \theta {)}^2+b^2{\sin }^2\theta$
$\Rightarrow P{G}^2={\left(\frac{a^2\cos \theta -(a^2-b^2)\cos \theta }{a}\right)}^2+b^2{\sin }^2\theta$
$\Rightarrow P{G}^2={\left(\frac{b^2\cos \theta }{a}\right)}^2+b^2{\sin }^2\theta$
$\Rightarrow P{G}^2=\frac{b^2}{a^2}(b^2{\cos }^2\theta +a^2{\sin }^2\theta )$

Similarly,
$\Rightarrow P{g}^2=\frac{a^2}{b^2}(b^2{\cos }^2\theta +a^2{\sin }^2\theta )$

$\therefore \frac{P{G}^2}{P{g}^2}=\frac{b^4}{a^4}$
$\Rightarrow \frac{PG}{Pg}=\frac{b^2}{a^2}$