Let P( x _ 0 , y _ 0 ) be the point on the hyperbola 3 x ^ 2 - 4 y...

Given,

Equation of hyperbola, $3{x}^{2} - 4{y}^{2} = 36$

And nearest line $3x + 2y = 1$

Now slope of the given line will be, $m = - \dfrac{3}{2}$

Now slope of tangent which will be parallel to given line of hyperbola $3{x}^{2} - 4{y}^{2} = 36$ is given by,

$m = \dfrac{3sec\theta}{\sqrt{12} \cdot tan\theta}$

Now putting the value of $m = \dfrac{- 3}{2}$ we get,

$⇒ \dfrac{3}{\sqrt{12}} \times \dfrac{1}{sin\theta} = \dfrac{- 3}{2}$

$⇒ sin\theta = - \dfrac{1}{\sqrt{3}}$

So, point will be \left(\sqrt{12}sec\theta, 3tan\theta\right)$</p><p>$= \left(\sqrt{12} \cdot \dfrac{\sqrt{3}}{\sqrt{2}}, - 3 \times \dfrac{1}{\sqrt{2}}\right)

$= \left(\dfrac{6}{\sqrt{2}}, \dfrac{- 3}{\sqrt{2}}\right)$,

Hence, $\sqrt{2}\left({y}_{0} - {x}_{0}\right) = \sqrt{2}\left(\dfrac{- 3}{\sqrt{2}} - \dfrac{6}{\sqrt{2}}\right) = - 9$