UPSC NDA I 2019 Mathematics Solved Paper

(B)
The minimum distance will occur along the normal to the parabola passing through (4,2) .

Normal to parabola at point Q( am‌2 ,−2 am) where a = 2 is
y = mx − 2.2.m − 2.m‌3 …. (i)
this normal passes through(4, 2) , hence
2 = 4m − 4m − 2m‌3
⇒ m‌3 = −1
⇒ m = −1. Hence point θ (2,(−1)‌2 ,−2∙2∙(−1))
⇒ θ (2,4)
Hence PQ=2√2