#### Need solution for RD Sharma maths class 12 chapter Maxima and Minima exercise Multiple Choice question, question 15.

Hint: Using Distance Formula, calculate the value of variables.

Given:  $y^2=2x$

Solution:

Let the required point be (x,y) which is nearest to (2,1).

$y^2=2x$

$\Rightarrow x=\frac{y^2}{4}$            …(i)

Now,

$d=\sqrt{(x-2)^2+(y-1)^2}$

Squaring on both sides, we get

\begin{aligned} &d^{2}=(x-2)^{2}+(y-1)^{2} \\ &d^{2}=\left(\frac{y^{2}}{4}-2\right)^{2}+(y-1)^{2} \\ &d^{2}=\frac{y^{4}}{16}+4-y^{2}+y^{2}+1-2 y \end{aligned}

Now,

$Z=d^2=\frac{y^2}{16}+5-2y$

$\frac{dZ}{dy}=\frac{y^3}{4}-2$                                                         …(ii)

For extrema$\frac{dZ}{dy}=0$

$\Rightarrow \frac{y^3}{4}-2=0$

$\Rightarrow \frac{y^3}{4}=2$

$\Rightarrow y^3=8$

$\Rightarrow y=2$

Substitute the value in Equation (i),

$x=\frac{2^2}{4}=1$

Now, differentiating (ii) w.r.t  y

\begin{aligned} &\frac{d^{2} Z}{d y^{2}}=\frac{3 y^{2}}{4} \\ &\frac{d^{2} Z}{d y^{2}}=\frac{3(2)^{2}}{4}=3>0 \end{aligned}

The nearest point is (1, 2).