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

Answer: (A)

Let $d$ be the distance of the point $(x,y)$ on $x^2=2y$ from the point $(0,5)$, then
$d=\sqrt{(x-0{)}^2+(y-5{)}^2}=\sqrt{2y+(y-5{)}^2}$.......(i)
$=\sqrt{2y+(y-5{)}^2}$ (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 $(y-4{)}^2=0$ i.e., when $y=4$
When $y=4$, then $x^2=2\times 4$
$\Rightarrow x=\pm \sqrt{8}=\pm 2\sqrt{2}$
$\therefore$ The points $(2\sqrt{2},4)$ and $(-2\sqrt{2},4)$ on the given curve are nearest to the point $(0,5)$. So, (a) is the correct option.