Question

The line $lx+my=n$ is a normal to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$ if

• A. $\frac{a^2}{l^2}+\frac{{(a^2-b^2)}^2}{n^2}=\frac{b^2}{m^2}$
• B. $\frac{a^2}{l^2}+\frac{b^2}{m^2}=\frac{{(a^2-b^2)}^2}{n^2}$
• C. $\frac{{(a^2-b^2)}^2}{n^2}+\frac{b^2}{m^2}=\frac{a^2}{l^2}$
• D. none of these

Answer 1

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

Normals of slope m to the ellipse are given by

$y=mx\pm \frac{m(a^2-b^2)}{\sqrt{a^2+b^2{m}^2}}$

so for $y=mx+c$

$c=\pm \frac{m(a^2-b^2)}{\sqrt{a^2+b^2{m}^2}}$

$c^2=\frac{m^2(a^2-b^2{)}^2}{a^2+b^2{m}^2}$

for $lx+my+n=0$

$y=-\frac{1}{m}x-\frac{n}{m}$

$c=-n/m$

slope$=-l/m$

$c^2=\frac{m^2(a^2-b^2{)}^2}{a^2+b^2{m}^2}$

$\frac{n^2}{m^2}=\frac{\frac{l^2}{m^2}{(a^2-b^2)}^2}{a^2+b^2\frac{l^2}{m^2}}$

$\frac{a^2{m}^2+b^2{l}^2}{l^2{m}^2}=\frac{(a^2-b^2{)}^2}{n^2}$

$\frac{a^2}{l^2}+\frac{b^2}{m^2}=\frac{(a^2-b^2{)}^2}{n^2}$.