If the point P on the curve, 4 x ^ 2 + 5 y ^ 2 = 20 is farthest...

Give ellipse

$\dfrac{{x}^{2}}{5} + \dfrac{{y}^{2}}{4} = 1 . . . \left(1\right)$

Let $P\left(\sqrt{5}cos\theta, 2sin\theta\right)$

Ellipse (1)

Given $Q\left(0, - 4\right)$

$P{Q}^{2} = {\left(\sqrt{{\left(\sqrt{5}cos\theta\right)}^{2} + {\left(2sin\theta\right)}^{2}}\right)}^{2}$

$⇒ P{Q}^{2} = 5\cos^{2}\theta + 4\sin^{2}\theta + 16\cos^{2}\theta + 16$

$⇒ P{Q}^{2} = 4 + \cos^{2}\theta + 16sin\theta + 16$

$= 1 - \sin^{2}\theta + 16sin\theta$

$⇒ P{Q}^{2} = 17 - \sin^{2}\theta + 16sin\theta$

$⇒ P{Q}^{2} = 17 - \left(\sin^{2}\theta - 16sin\theta\right)$

$⇒ P{Q}^{2} = 21\left({\left(sin\theta - 8\right)}^{2} - 64\right)$

$⇒ P{Q}^{2} = 85 - ( sin\theta - 8{)}^{2} = 85 - 49$

$MaxP{Q}^{2} = 36$ ($∵$ maximum value of $sin\theta = 1$)