JEE Main 2020 (Online) 5th September Morning ShiftQuestion 1
If the point $P$ on the curve, $4x^2+5y^2=20$ is farthest from the point $Q(0,-4)$, then $PQ^2$ is equal to :
Differentiation and its Application
Solution
Concept. Write $PQ^2$ in terms of $y$ alone, using the ellipse to remove $x^2$. Formulas used. (1) $4x^2+5y^2=20$ $\Rightarrow$ $x^2=5-\frac54y^2$, $\ -2\le y\le2$Step 1 - Distance. By (1), $$PQ^2=x^2+(y+4)^2=21+8y-\frac{y^2}{4}.$$ Step 2 - Maximise. The derivative $8-\frac y2>0$ on $[-2,2]$, so the maximum is at $y=2$, i.e. $P=(0,2)$.Step 3 - Value. $PQ^2=21+16-1=36$. Answer: (A)