Question

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

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

Answer 1

Answer: (B), (D)

The correct options are
B $(2\sqrt{2},4).$
D $(-2\sqrt{2},4).$

Given, $x^2=2y$

An point on the curve is $(x,\frac{x^2}{2})$

$D=\sqrt{x^2+(5-\frac{x^2}{2}{)}^2}$

$D^2=S=x^2+(5-\frac{x^2}{2}{)}^2$

$\frac{dS}{dx}=2x+2(5-x^{2}2).(-x)$
$=2x-10x+x^3$
$=x^3-8x$
$=0$
$x=0$ and $x=\pm 2\sqrt{2}$

Hence, the required points are

$\Rightarrow (x,\frac{x^2}{2})$

$\Rightarrow (-2\sqrt{2},4),(2\sqrt{2},4)$.