Question

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

• A. $(2\sqrt{2},4)$
• B. $(4,8)$
• C. $(\sqrt{2},1)$
• D. $(2,2)$

Answer 1

The correct option is A $(2\sqrt{2},4)$

Let $P(x,y)$ be a point on the curve $x^2=2y$
Let $S$ be the distance between $P$ and $(0,5)$.
$S^2=(x-0{)}^2+(y-5{)}^2$
$S^2=2y+(y-5{)}^2$
$2S\frac{dS}{dy}=2+2(y-5)$
For maxima or minima,
$\frac{dS}{dy}=0$
$\Rightarrow y=4$
$\frac{d^{2}S}{d{y}^2}>0$ at $y=4$
$\Rightarrow x=\pm 2\sqrt{2}$
Required point is $(2\sqrt{2},4)$