Question

Find the point on the curve ${x^2} = 2y$ which is closest to the point $\left( {0,5} \right)$ .
A. $\left( { \pm 2\sqrt 2 ,4} \right)$
B. $\left( { \pm 4\sqrt 2 ,16} \right)$
C. $\left( { \pm \sqrt 2 ,1} \right)$
D. None of these

Answer 1

In the above given question, we are given an equation of a curve, that is ${x^2} = 2y$ . We have to find a point on the given curve which is closest to another point $\left( {0,5} \right)$. In order to approach the solution we have to suppose the required point on the curve which is closest to the point $\left( {0,5} \right)$ to be as $\left( {h,k} \right)$ . After that, we can apply the distance formula of 2 dimensional coordinate geometry. Then we can find the value of the point $\left( {h,k} \right)$ for the minimum distance using first and second derivatives.

Complete step by step answer:
Given equation of the curve is,
$\Rightarrow {x^2} = 2y$
We have to find the point on the given curve which is closest to the point $\left( {0,5} \right)$. Let the point on the curve ${x^2} = 2y$ which is closest to the point $\left( {0,5} \right)$ be $\left( {h,k} \right)$. Then,
$\Rightarrow {h^2} = 2k$
Then, using the distance formula, the distance between the two points $\left( {h,k} \right)$ and $\left( {0,5} \right)$ can be written as,
$\Rightarrow d = \sqrt {{{\left( {h - 0} \right)}^2} + {{\left( {k - 5} \right)}^2}}$

Let, $D = {d^2}$ then we can write the above equation as,
$\Rightarrow D = {d^2} = {\left( {h - 0} \right)^2} + {\left( {k - 5} \right)^2}$
That gives us ,
$\Rightarrow D = {h^2} + {k^2} + 25 - 10k$
Substituting ${h^2} = 2k$ in above equation, we get
$\Rightarrow D = 2k - 10k + {k^2} + 25$
$\Rightarrow D = - 8k + {k^2} + 25$
Differentiating both sides with respect to $k$ , we get
$\Rightarrow \dfrac{{dD}}{{dk}} = 2k - 8$
For a maximum or minimum value, the first derivative must be zero.
Hence, putting $\dfrac{{dD}}{{dk}} = 0$ we get
$\Rightarrow \dfrac{{dD}}{{dk}} = 2k - 8 = 0$
That gives us,
$\Rightarrow k = 4$

Now the second derivative of the curve can be written as,
$\Rightarrow \dfrac{{{d^2}D}}{{d{k^2}}} = 2 > 0$
Hence, the curve has a minimum distance from $\left( {0,5} \right)$ when $k = 4$ .
Also, when $k = 4$ , we have
$\Rightarrow {h^2} = 2\left( 4 \right)$
That gives us,
$\Rightarrow {h^2} = 8$
Hence,
$\Rightarrow h = \pm \sqrt 8$
$\therefore h = \pm 2\sqrt 2$
Hence, the required point on the curve is $\left( {h,k} \right) = \left( { \pm 2\sqrt 2 ,4} \right)$. Therefore, the point on the curve ${x^2} = 2y$ which is closest to the point $\left( {0,5} \right)$ is $\left( { \pm 2\sqrt 2 ,4} \right)$.

Therefore, the correct option is A.

Note: The given equation of the curve, that is ${x^2} = 2y$ , is the equation of a parabola which is lying in the first and second quadrant. Since, from the equation ${x^2} = 2y$ we can say that on the y-axis the curve is always in the positive direction, while on the x-axis the curve can have both the positive as well as negative direction, i.e. values. The graph of the curve will be a $\cup$ shaped parabola with the vertex at origin.