Question

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

• A. $(4,8)$
• B. $(1,1/2)$
• C. $(2\sqrt{2},4)$
• D. None of these

Answer 1

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

Let $(h,k)$ lie on the curve $x^2=2y$ which is nearest to the point $(0,5)$

Since $(h,k)$ lie on the curve $x^2=2y$

$\Rightarrow (h,k)$ will satisfy the equation of curve $x^2=2y$

$\Rightarrow$ Putting $x=h$ and $y=k$ in equation we get

$h^2=2k$ ......$\left(1\right)$

We need to minimize the distance of a point $(h,k)$ from $(0,5)$

Let $D$ be the distance between $(h,k)$ and $(0,5)$

$D=\sqrt{h^2+{(5-k)}^2}$

From $\left(1\right)$ we have $h^2=2k$

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

Differentiating w.r.t $k$ we get

$\frac{dD}{dk}=\frac{1}{2\sqrt{2k+{(5-k)}^2}}\times \frac{d}{dk}(2k+{(5-k)}^2)$

$=\frac{1}{2\sqrt{2k+{(5-k)}^2}}\times [2+2(5-k)\frac{d}{dk}(5-k)]$

$=\frac{1}{2\sqrt{2k+{(5-k)}^2}}\times [2-2(5-k)]$

$=\frac{-4+k}{\sqrt{2k+{(5-k)}^2}}$

Put $\frac{dD}{dk}=0$

$\Rightarrow \frac{-4+k}{\sqrt{2k+{(5-k)}^2}}=0$

$\Rightarrow -4+k=0$

$\therefore k=4$ is a point of minima.

$D$ is minimum when $k=4$

We have $h^2=2k=2\times 4=8$

$\therefore h=2\sqrt{2}$

Hence the required point is $(h,k)=(2\sqrt{2},4)$