The point on the ellipse x^ 2 6 + y^ 2 3 = 1 which is nearest to...

Shortest distance always lies along the common normal
$⇒$ Slope of common normal is $1$
Now, the equation of the normal at the point \left(acos \theta, bsin \theta\right)on\dfrac{x2}{a2} + \dfrac{y2}{b2} = 1$is<br />$\dfrac{ax}{cos\theta} - \dfrac{by}{sin\theta} = a^{2} - b^{2},$where slope is$\dfrac{a}{b}tan \theta$<br />$⇒ \dfrac{\sqrt{6}}{\sqrt{3}}tan \theta = 1 ⇒ tan \theta = \dfrac{1}{\sqrt{2}}$<br />$⇒ sin \theta = \dfrac{1}{\sqrt{3}}$and$cos \theta = \dfrac{\sqrt{2}}{\sqrt{3}}$<br />or$sin \theta = - \dfrac{1}{3}, cos \theta = - \dfrac{2}{3}$~(rejected because it gives farthest point)<br />$⇒Nearest point is \left(\sqrt{6}\left(\dfrac{\sqrt{2}}{\sqrt{3}}\right), \sqrt{3}\left(\dfrac{1}{\sqrt{3}}\right)\right) = \left(2,1\right)