Which of the following points lies on the locus of the foot of perpendicular drawn upon any
tangent to the ellipse, $\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$ from any of its foci ?
Correct Option: 1
Let foot of perpendicular is $(\mathrm{h}, \mathrm{k})$
$\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$ (Given)
$a=2, b=\sqrt{2}, e=\sqrt{1-\frac{2}{4}}=\frac{1}{\sqrt{2}}$
$\therefore$ Focus $(\mathrm{ae}, 0)=(\sqrt{2}, 0)$
Equation of tangent
$y=m x+\sqrt{a^{2} m^{2}+b^{2}}$
$y=m x+\sqrt{4 m^{2}+2}$
Passes throguh $(h, k)$
$(\mathrm{k}-\mathrm{mh})^{2}=4 \mathrm{~m}^{2}+2$ .................(1)
line perpendicular to tangent will have slope
$-\frac{1}{m}$
$y-0=-\frac{1}{m}(x-\sqrt{2})$
$m y=-x+\sqrt{2}$
$(h+m k)^{2}=2$ .............(2)
Add equaiton (1) and (2)
$\mathrm{k}^{2}\left(1+\mathrm{m}^{2}\right)+\mathrm{h}^{2}\left(1+\mathrm{m}^{2}\right)=4\left(1+\mathrm{m}^{2}\right)$
$\mathrm{h}^{2}+\mathrm{k}^{2}=4$
$\mathrm{x}^{2}+\mathrm{y}^{2}=4$ (Auxilary circle)
$\therefore(-1, \sqrt{3})$ lies on the locus.
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