Question

The point $(0,5)$ is closer to the curve $x^2=2y$ at:

• A. $(2\sqrt{2},0)$
• B. $(0,0)$
• C. $(2,2)$
• D. none of these

Answer 1

Answer: (D)

none of these

Let any point on the curve be $(h,k)$.
Then, $h^2=2k$
$\therefore$ Distance $D=\sqrt{h^2+{(k-5)}^2}$
$\Rightarrow D=\sqrt{2k+{(k-5)}^2}$
$\Rightarrow \frac{dD}{dk}=\frac{1}{2\sqrt{2k+{(k-5)}^2}}\times \{2+2(k-5)\}=0$ say
$k=4$
Then, point will be $(\pm 2\sqrt{2},4)$