Question

The point, which is at the shortest distance from the line $x+y=7$ and lying on an ellipse $x^2+2y^2=6$ , has coordinates $(a,b)$ then the value of $\frac{a}{b}$ is

Answer 1

Answer: 2

Equation of the ellipe is $\frac{x^2}{6}+\frac{y^2}{3}=1$
The tangent at the point of shortest distance from the line $x+y=7$ should be 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
$\frac{xcos\theta }{\sqrt{6}}+\frac{ysin\theta }{\sqrt{3}}=1.$ It is parallel to $x+y=7$
$\because$ Equation of the tangent at $P(acos\theta ,bsin\theta )$ on ellipse is $\frac{xcos\theta }{a}+\frac{ysin\theta }{b}=1$
$\Rightarrow \frac{cos\theta }{\sqrt{6}}=\frac{sin\theta }{\sqrt{3}}$
$\Rightarrow \frac{cos\theta }{\sqrt{2}}=\frac{sin\theta }{1}$ $\Rightarrow sin\theta =\frac{1}{\sqrt{3}},cos\theta =\sqrt{\frac{2}{3}}$
The required point is $(2,1)$
Hence, $(a,b)=(2,1)$
therefore, $\frac{a}{b}=2$ .