Question

The point on the curve $x^2=2y$ which is nearest to the point (0, 5) is

(A) $(2\sqrt{2},4$) (B) $(2\sqrt{2},0$) (C) (0, 0) (D) (2, 2)

Answer 1

Let d be the distance of the point (x,y) on $x^2=2y$ from the point (0,5) then
$d=\sqrt{(x-o{)}^2+(y-5{)}^2}=\sqrt{x^2+(y-5{)}^2}\dots \left(i\right)$
$=\sqrt{2y+(y-5{)}^2}$ (Putting $x^2=2y$)
$=\sqrt{y^2-8y+25}=\sqrt{y^2-8y+4^2+9}=\sqrt{(y-4{)}^2+9}$
d is least when $(2\sqrt{2},4)$ and $(-2\sqrt{2},4)$ on the given curve are nearest to the point (0,5), so, (a) is the correct option.