Question

The locus of the foot of the perpendicular from the centre of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ on any tangent is given by $(x^2+y^2)=l{x}^2+m{y}^2$,where

• A. $l=a^2,m=b^2$
• B. $l=-a^2,b=-b^2$
• C. $l=-a^2,m=-b^2$
• D. $l=a^2,b=-b^2$

Answer 1

Answer: (A)

Equation of any tangent to the given ellipse is
$y=mx\pm \sqrt{a^2{m}^2+b^2}$
$y-mx=\pm \sqrt{a^2{m}^2+b^2}...\left(1\right)$
Equation of perpendicular line is $my+x=\lambda$
It passes through the centre $\left(0,0\right)$
$\therefore \lambda =0$
$my+x=0...\left(2\right)$
On squaring and adding $\left(1\right)$ and $\left(2\right)$, we get
$y^2+m^2{x}^2+m^2{y}^2+x^2=a^2{m}^2+b^2$
$\left(1+m^2\right)\left(x^2+y^2\right)=a^2{m}^2+b^2$
$\Rightarrow \left(1+\frac{x^2}{y^2}\right)\left(x^2+y^2\right)=\frac{a^2{x}^2}{y^2}+b^2$
$\left[from\left(2\right)\right]$
$\Rightarrow {\left(x^2+y^2\right)}^2=a^2{x}^2+b^2{y}^2$
But ${\left(x^2+y^2\right)}^2=l{x}^2+m{y}^2$
$l=a^2,m=b^2$