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. $(2,2)$
• E. $(2\sqrt{2},2)$

Answer 1

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

Point on a parabola can be taken to be $(2at,a{t}^2)$.

Here $a=\frac{1}{2}$

So the point is $(t,\frac{t^2}{2})$

Distance of this point from $(0,5)$ will be $\sqrt{t^2+{(\frac{t^2}{2}-5)}^2}$

Lets minnimise $D^2$ and we will get minimum distance automatically.

First derivative of $D^2$ is $t^3-8t$

Putting this to zero we get $t=0$ and $t=\pm 2\sqrt{2}$

Now let's do double derivative, it comes out to be $3t^2-8$.

So, we can see that for $t=0$, distance is going to be maximum.

So, distance is minimum from $(2\sqrt{2},4)$.