The point, which is at the shortest distance from the line x + y =...

Equation of the ellipe is $\dfrac{{x}^{2}}{6} + \dfrac{{y}^{2}}{3} = 1$

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

$\dfrac{xcos \theta}{\sqrt{6}} + \dfrac{ysin \theta}{\sqrt{3}} = 1 .$ It is parallel to $x + y = 7$

$∵$ Equation of the tangent at $P( acos\theta, bsin\theta )$ on ellipse is $\dfrac{xcos\theta}{a} + \dfrac{ysin\theta}{b} = 1$

$⇒ \dfrac{cos \theta}{\sqrt{6}} = \dfrac{sin \theta}{\sqrt{3}}$

$⇒ \dfrac{cos \theta}{\sqrt{2}} = \dfrac{sin \theta}{1}$$⇒ sin\theta = \dfrac{1}{\sqrt{3}} , cos\theta = \sqrt{\dfrac{2}{3}}$

The required point is $( 2, 1 )$

Hence, $( a, b ) = ( 2, 1 )$

therefore, $\dfrac{a}{b} = 2$.