On the ellipse x ^ 2 8 + y ^ 2 4 = 1 , let P be a point in the...

We have an ellipse $\dfrac{{x}^{2}}{8} + \dfrac{{y}^{2}}{4} = 1$, we have a general point $P ( \sqrt{8}cos\theta, 2sin\theta )$

Equation of tangent at point $p$

$\dfrac{x}{\sqrt{8}}cos\theta + \dfrac{y}{2}sin\theta - 1 = 0$

Slope $= \dfrac{- 1}{\sqrt{2}}cot\theta$

Given that, tangent at $P$ is perpendicular to the line $x + 2y = 0$.

So, product of their slopes are perpendicular $⇒ \dfrac{- 1}{\sqrt{2}}cot\theta \times \dfrac{- 1}{2} = - 1$

$⇒ \dfrac{- 1}{\sqrt{2}}cot\theta = 2$

$⇒ cot\theta = - 2\sqrt{2}$

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

So, point $P\left(\dfrac{- 8}{3}, \dfrac{2}{3}\right)$

$A = \dfrac{1}{2} \times 2ae \times \dfrac{2}{3}$

Where, $e = \sqrt{\dfrac{{a}^{2} - {b}^{2}}{{a}^{2}}} = \sqrt{\dfrac{8 - 4}{8}} = \dfrac{1}{\sqrt{2}}$

$∴ A = \dfrac{1}{2} \times 2 \times 2\sqrt{2} \times \dfrac{1}{\sqrt{2}} \times \dfrac{2}{3} = \dfrac{4}{3}$So, \left(5 - {e}^{2}\right)A = \left(5 - \dfrac{1}{2}\right) \times \dfrac{4}{3} = 6