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

Hint: 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.