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Q.
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 $(x^2+y^2) = lx^2+my^2$,where
Conic Sections
Solution:
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}}\quad...\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 \quad...\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}$
$\quad\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} = lx^{2}+my^{2} $
$ l=a^{2}, m =b^{2} $