Question

The point (0,5) would be closest to the curve ${{x}^{2}}=2y$ at
a) $\left( 2\sqrt{2},0 \right)$
b) (0,0)
c) (2,2)
d) none of these

Answer 1

Hint: The point will be closest to the curve when the distance between the curve and the point will be the smallest. Therefore, we can minimise the distance function from the curve to the point to find the solution.
complete step-by-step solution -
We are required to find the point when the point (0,5) will be closest to the curve ${{x}^{2}}=2y$ i.e. the distance between the point and the curve will be minimum.
Therefore, we should construct a function for the distance between the point (0,5) and an arbitrary point on the curve.
The distance between two points $\left( {{a}_{1}},{{b}_{1}} \right)$ and $\left( {{a}_{2}},{{b}_{2}} \right)$ in Euclidean geometry is given by $\sqrt{{{\left( {{a}_{1}}-{{a}_{2}} \right)}^{2}}+{{\left( {{b}_{1}}-{{b}_{2}} \right)}^{2}}}.......(1.1)$
Let $\left( x,y \right)$ be any point on the given curve i.e. its coordinates satisfy the equation ${{x}^{2}}=2y......(1.2)$. Then its distance from the point (0,5) would be given from equation (1.1) as
$f(x,y)=\sqrt{{{\left( x-0 \right)}^{2}}+{{\left( y-5 \right)}^{2}}}=\sqrt{{{x}^{2}}+{{(y-5)}^{2}}}=\sqrt{2y+{{(y-5)}^{2}}}....(\text{from equation (1}\text{.2)})$
Now, to find the minimum distance, we should minimise f with respect to y (as f is represented by the single variable y in the above equation). However, we can see that f(x) would be minimum when the term inside the square-root is minimum (as the sum of squares is positive and the square root of a higher number has a higher value). Thus, we can also minimise the function
$g(y)=2y+{{(y-5)}^{2}}$
The conditions of minima of a function f at $y={{y}_{0}}$ can be stated as:
${{\left[ \dfrac{df}{dy} \right]}_{y={{y}_{0}}}}=0$ i.e. its derivative vanishes at $y={{y}_{0}}$
${{\left[ \dfrac{{{d}^{2}}f}{d{{y}^{2}}} \right]}_{y={{y}_{0}}}}>0$
So, in this case, if the distance function attains a minimum at $y={{y}_{0}}$, then g(y) will also attend a minimum and thus,
$\begin{align}
  & {{\left[ \dfrac{dg}{dy} \right]}_{y={{y}_{0}}}}={{\left[ \dfrac{d(2y+{{(y-5)}^{2}})}{dy} \right]}_{y={{y}_{0}}}}=0 \\
 & \Rightarrow {{\left[ 2+2(y-5) \right]}_{y={{y}_{0}}}}=0\Rightarrow {{y}_{0}}-5=-1\Rightarrow {{y}_{0}}=4 \\
\end{align}$

Now, we check the second condition at the obtained point
${{\left[ \dfrac{{{d}^{2}}g}{d{{y}^{2}}} \right]}_{y= 4}}={{\left[ \dfrac{d}{dy}\left( \dfrac{dg}{dy} \right) \right]}_{y=4}}=\dfrac{d(2+2(y-5))}{dy}=2>0$
Thus the second condition for minima is also satisfied.
Thus, the distance of the function is minimum at the y coordinate 4. However, no option in the question has y coordinate 4, thus the correct option should be none of the above.

Note: The answer could also have been found by minimising the whole function instead of the term inside the square root. However, that would have resulted in more steps and resulted in more calculations without changing the result.