Question

If the straight line $lx+my+n=0$ is a normal to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, then $\frac{a^2}{l^2}+\frac{b^2}{m^2}$ is equal to

• A. $\frac{(a^2-b^2{)}^2}{n}$
• B. $\frac{(a^2-b^2{)}^2}{n^2}$
• C. $\frac{a^2-b^2}{n}$
• D. $\frac{(a^2+b^2{)}^2}{n^2}$

Answer 1

The correct option is B $\frac{(a^2-b^2{)}^2}{n^2}$

We know that, equation of any normal is $ax\sec \theta -bycosec\theta =a^2-b^2...\left(i\right)$
Since, it is given that straight line
$lx+my+n=0....\left(ii\right)$
is a normal to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
So, Eqs. (i) and (ii) represent the same line
$\therefore \frac{a\sec \theta }{bcosec\theta }m=\frac{a^2-b^2}{-n}$
$\Rightarrow \cos \theta =\frac{-an}{(a^2-b^2)}$
and $\sin \theta =\frac{bn}{m(a^2-b^2)}$
$\because {\cos }^2\theta +{\sin }^2\theta =1$
$\therefore {\left(\frac{-an}{l(a^2-b^2)}\right)}^2+{\left(\frac{bn}{m(a^2-b^2)}\right)}^2=1$
$\Rightarrow \frac{(a^2-b^2{)}^2}{n^2}=\frac{a^2}{l^2}+\frac{b^2}{m^2}$.