If the tangent at a point P on the parabola y ^ 2 = 3x is parallel...

Given equation of parabola is

${y}^{2} = 3x$

$⇒ 2y\dfrac{dy}{dx} = 3$

$⇒ {\left(\dfrac{dy}{dx}\right)}_{slopeoftangentatP} = \dfrac{3}{2y}$

Tangent at $P\left({x}_{1}, {y}_{1}\right)$ is parallel to the line $x + 2y = 1$, so slope of tangent is

$m = - \dfrac{1}{2}$

$⇒ \dfrac{3}{2{y}_{1}} = - \dfrac{1}{2}$

$⇒ {y}_{1} = - 3$

So, coordinates of $P ≡ \left(3, - 3\right)$.

Given equation of ellipse is

$\dfrac{{x}^{2}}{4} + \dfrac{{y}^{2}}{1} = 1$

$⇒ {x}^{2} + 4{y}^{2} = 4$

$⇒ 2x + 8y\left(\dfrac{dy}{dx}\right) = 0$

$⇒ {\left(\dfrac{dy}{dx}\right)}_{slopeoftangent} = - \dfrac{x}{4y}$

Also, tangent to ellipse re perpendicular to line $x - y = 2$, so slope of tangent

$m = - 1$

$⇒ - \dfrac{x}{4y} = - 1$

$⇒ x = 4y$

Putting in equation of ellipse, we get

$\dfrac{16{y}^{2}}{4} + \dfrac{{y}^{2}}{1} = 1$

$⇒ y = \pm \dfrac{1}{\sqrt{5}}$

So,

$Q\left(\dfrac{4}{\sqrt{3}}, \dfrac{1}{\sqrt{5}}\right), R\left(- \dfrac{4}{\sqrt{5}}, \dfrac{- 1}{\sqrt{5}}\right)$

Area of $∆ PQR$ is

= \dfrac{1}{2}\left|\begin{cases}3 & - 3 & 1 \\ \dfrac{4}{\sqrt{5}} & \dfrac{1}{\sqrt{5}} & 1 \\ - \dfrac{4}{\sqrt{5}} & - \dfrac{1}{\sqrt{5}} & 1\end{cases}\right|

$= \dfrac{1}{2}\left[3\left(\dfrac{2}{\sqrt{5}}\right) + 3\left(\dfrac{8}{\sqrt{5}}\right) + 0\right]$

$= \dfrac{30}{2\sqrt{5}} = 3\sqrt{5}$ sq. units