Question

$\mathrm{The}\mathrm{straight}\mathrm{line}\mathrm{lx}+\mathrm{my}+n=0\mathrm{touches}\mathrm{the}\mathrm{ellipse}\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\mathrm{if}$

• A. $a^2{l}^2+b^2{m}^2=n^2$
• B. $a^2{l}^2+b^2{m}^2=n$
• C. $\mathrm{al}+\mathrm{bm}=n$
• D. $a^2{l}^2+b^2{m}^2=0$

Answer 1

The correct option is A $a^2{l}^2+b^2{m}^2=n^2$

The equaton of the line which touches the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is
$y=\mathrm{px}\pm \sqrt{a^2{p}^2+b^2}\Rightarrow \mathrm{px}-y\pm \sqrt{a^2{p}^2+b^2}=0$
comparing the above equation to the given qeuation, $\mathrm{lx}+\mathrm{my}+n=0$
we have
$\frac{p}{l}=\frac{-1}{m}=\frac{\pm \sqrt{a^2{p}^2+b^2}}{n}$
Eliminating p from above equations
$a^2{\left(-\frac{l}{m}\right)}^2+b^2=\frac{n^2}{m^2}{a}^2{l}^2+b^2{m}^2=n^2$