The point on the curve
Question:

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

(A) $(2 \sqrt{2}, 4)$

(B) $(2 \sqrt{2}, 0)$

(C) $(0,0)$

(D) $(2,2)$

Solution:

The given curve is $x^{2}=2 y$.

For each value of $x$, the position of the point will be $\left(x, \frac{x^{2}}{2}\right)$.

Let $\mathrm{P}\left(x, \frac{x^{2}}{2}\right)$ and $\mathrm{A}(0,5)$ are the given points.

Now distance between the points $\mathrm{P}$ and $\mathrm{A}$ is given by,

$\begin{aligned} \mathrm{PA} &=\sqrt{(x-0)^{2}+\left(\frac{x^{2}}{2}-5\right)^{2}} \\ \Rightarrow \mathrm{PA}^{2} &=(x-0)^{2}+\left(\frac{x^{2}}{2}-5\right)^{2} \\ \Rightarrow \mathrm{PA}^{2} &=x^{2}+\frac{x^{4}}{4}+25-5 x^{2} \\ \Rightarrow \mathrm{PA}^{2} &=\frac{x^{4}}{4}-4 x^{2}+25 \\ \Rightarrow \mathrm{PA}^{2} &=y^{2}-8 y+25 \quad\left(\text { as }, x^{2}=2 y\right) \end{aligned}$

Let us denote PA $^{2}$ by Z. Then,

$\mathrm{Z}=y^{2}-8 y+25$

Differentiating both sides with respect to y, we get

$\frac{d Z}{d y}=2 y-8$

For maxima or minima, we have

$\frac{d Z}{d y}=0$

$\Rightarrow 2 y-8=0$

$\Rightarrow y=4$

$\frac{d^{2} Z}{d y^{2}}=2$

Now,

$\left[\frac{d^{2} Z}{d y^{2}}\right]_{y=4}=2>0$

Now, $x^{2}=2 y$

$\Rightarrow x^{2}=2 \times 4$

$\Rightarrow x^{2}=8$

$\Rightarrow x=2 \sqrt{2}$ or $x=-2 \sqrt{2}$

So, $Z$ is minimum at $(2 \sqrt{2}, 4)$ or $(-2 \sqrt{2}, 4)$.

Or, $\mathrm{PA}^{2}$ is minimum at $(2 \sqrt{2}, 4)$ or $(-2 \sqrt{2}, 4)$.

 

Or, PA is minimum at $(2 \sqrt{2}, 4)$ or $(-2 \sqrt{2}, 4)$.

So, distance between the points $P\left(x, \frac{x^{2}}{2}\right)$ and $A(0,5)$ is minimum at $(2 \sqrt{2}, 4)$ or $(-2 \sqrt{2}, 4)$.

So, the correct answer is A.

 

 

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