Question

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

• A. $(2,2),(-2,2)$
• B. $(2\sqrt{2},4),(-2\sqrt{2},4)$
• C. $(\sqrt{6},3),(-\sqrt{6},3)$
• D. $(2\sqrt{3},6),(-2\sqrt{3},6)$

Answer 1

Answer: (B)

Let $(h,k)$ be the point on the curve
$x^2=2y$ $ie,$ $h^2=2k$ ... (i)
Let $D$ be the distance between $(h,k)$ and $(0,5)$ .
$\therefore$ $D=\sqrt{h^2+{(k-5)}^2}$
$=\sqrt{2k{(k-5)}^2}$
On differentiating w.r.t. $k,$
we get $\frac{dD}{dk}=\frac{2+2(k-5)}{\sqrt{2k+{(k-5)}^2}}$
For minima, put $\frac{dD}{dk}=0$
$\Rightarrow$ $2+2(k-5)=0\Rightarrow k=4$
$\therefore$ From Eq. (i), we get $h^2=2\times 4$
$\Rightarrow$ $h=\pm 2\sqrt{2}$
Hence, closest point is $(\pm 2\sqrt{2},4)$ .