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

We have,

$x^{2} + 3y^{2} = 6$

$⇒ \dfrac{x2}{6} + \dfrac{y2}{2} = 1$

Centre of an ellipse is~\left(0, 0\right)$.</p><p>Let the foot of perpendicular be$( h, k )$.</p><p>Equation of tangent in slope form to the standard ellipse~$\dfrac{x2}{a2} + \dfrac{y2}{b2} = 1$~is</p><p>$y = mx \pm \sqrt{a^{2}m^{2} + b^{2}}$.</p><p>Hence, equation of tangent with slope$m$passing$( h, k )$is~</p><p>$y = mx \pm \sqrt{6m^{2} + 2}$</p><p>$⇒ k = mh \pm \sqrt{6m^{2} + 2}$, where$m = - \dfrac{h}{k}$.</p><p>$⇒ k = h\left(- \dfrac{h}{k}\right) \pm \sqrt{6m^{2} + 2}$</p><p>$⇒ \sqrt{\dfrac{6h^{2}}{k2} + 2} = \dfrac{h^{2} + k^{2}}{k}$</p><p>$⇒ 6h^{2} + 2k^{2} = \left(h^{2} + k^{2}\right)2$</p><p>So, required locus is</p><p>$6x^{2} + 2y^{2} = \left(x^{2} + y^{2}\right)2.