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 $\left(\right. a , b \left.\right)$ then the value of $\frac{a}{b}$ is

Answer 1

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 $\left(\right.\sqrt{6}cos \theta , \, \sqrt{3}sin ⁡ \theta \left.\right)$
Equation of the tangent is
$\frac{x cos \theta }{\sqrt{6}}+\frac{y sin ⁡ \theta }{\sqrt{3}}=1.$ It is parallel to $x+y=7$
$\because $ Equation of the tangent at $P \left(\right. a cos \theta , b sin \theta \left.\right)$ on ellipse is $\frac{x cos \theta }{a} + \frac{y sin \theta }{b} = 1$
$\Rightarrow \frac{cos \theta }{\sqrt{6}}=\frac{sin ⁡ \theta }{\sqrt{3}}$
$\Rightarrow \frac{cos \theta }{\sqrt{2}}=\frac{sin ⁡ \theta }{1}$ $\Rightarrow sin \theta = \frac{1}{\sqrt{3}} , cos \theta = \sqrt{\frac{2}{3}}$
The required point is $\left(\right.2, \, 1\left.\right)$
Hence, $\left(\right. a , b \left.\right) = \left(\right. 2 , 1 \left.\right)$
therefore, $\frac{a}{b} = 2$ .