The locus of the foot of perpendicular drawn from the centre of the...

The centre of the given ellipse $\left(0, 0\right)$
The general equation of a tangent to the ellipse

$\dfrac{{x}^{2}}{{a}^{2}} + \dfrac{{y}^{2}}{{b}^{2}} = 1$ is $y = mx \pm \sqrt{{a}^{2}{m}^{2} + {b}^{2}} . . . \left(1\right)$

Given ellipse : $\dfrac{{x}^{2}}{6} + \dfrac{{y}^{2}}{2} = 1$

A perpendicular line from the centre is $y = - \dfrac{x}{m} . . . \left(2\right)$

Eliminating $m$, from \left(1\right)$and \left(2\right)$

$y = \left(\dfrac{- x}{y}\right)x \pm \sqrt{{a}^{2}\left(\dfrac{{x}^{2}}{{y}^{2}}\right) + {b}^{2}}$

${y}^{2} = - {x}^{2} \pm y\sqrt{{a}^{2}\left(\dfrac{{x}^{2}}{{y}^{2}}\right) + {b}^{2}}$

$∴ {x}^{2} + {y}^{2} = \pm \sqrt{{a}^{2}{x}^{2} + {b}^{2}{y}^{2}}$

Squaring both sides,

${\left({x}^{2} + {y}^{2}\right)}^{2} = {a}^{2}{x}^{2} + {b}^{2}{y}^{2} = 6{x}^{2} + 2{y}^{2}$