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The locus of the foot of perpendicular drawn from the centre of the ellipse ${x^2} + 3{y^2} = 6$ on any tangent to it is
${\left( {{x^2} + {y^2}} \right)^2} = 6{x^2} + 2{y^2}$
$\;{\left( {{x^2} + {y^2}} \right)^2} = 6{x^2} - 2{y^2}$
$\;{\left( {{x^2} - {y^2}} \right)^2} = 6{x^2} + 2{y^2}$
$\;{\left( {{x^2} - {y^2}} \right)^2} = 6{x^2} - 2{y^2}$
Solution
$x^{2}+3 y^{2}=6$
$\Rightarrow \frac{x^{2}}{6}+\frac{y^{2}}{2}=1$
Now, we know that any tangent to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is given by:
$y=m x+\sqrt{a^{2} m^{2}+b^{2}}$
So, the equation of the tangent to the given ellipse is:-
$y=m x+\sqrt{6 m^{2}+2}$ ………..$(1)$
Now, equation of the line through $(0,0)$ and perpendicular to $(1)$ is:-
$y-0=\left(-\frac{1}{m}\right) x-0$
$\Rightarrow m=-\frac{x}{y}$ …..$(2)$
Using $( 2)$ in $(1),$ we have:
$y=\left(-\frac{x}{y}\right) x+\sqrt{6\left(\frac{x^{2}}{y^{2}}\right)+2}$
$\Rightarrow y^{2}=-x^{2}+\sqrt{6 x^{2}+2 y^{2}}$
$\Rightarrow\left(x^{2}+y^{2}\right)^{2}=6 x^{2}+2 y^{2}$
