Q. The point, at shortest distance from the line $x + y = 7$ and lying on an ellipse $x^{2} + 2y^{2} = 6$, has coordinates
A) $\left( \sqrt{2},\ \sqrt{2} \right)$
B) $\left( 0,\ \sqrt{3} \right)$
C) $(2,\ 1)$
D) $\left( \sqrt{5},\ \frac{1}{\sqrt{2}} \right)$
Explanation
(c)
The tangent at the point of shortest distance from the line $x + y = 7$ parallel to the given line
Any point on the given ellipse is $(\sqrt{6}\cos\theta,\ \sqrt{3}\sin\theta)$
Equation of the tangent is
$\frac{x\cos\theta}{\sqrt{6}} + \frac{y\sin\theta}{\sqrt{3}} = 1.$ It is parallel to $x + y = 7$
$\Rightarrow \frac{\cos\theta}{\sqrt{6}} = \frac{\sin\theta}{\sqrt{3}}$
$\Rightarrow \frac{\cos\theta}{\sqrt{2}} = \frac{\sin\theta}{1} = \frac{1}{\sqrt{3}}$
The required point is $(2,\ 1)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.