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

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

Answers (1)

Answer:  (1,2)

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).

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