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Please solve RD Sharma class 12 Chapter maxima and minima exercise 17.5 question 33 maths textbook solution.

Answers (1)

(-2, -8)

Hint: For maximum or minimum value of S, we must have \frac{dS}{dt}=0

Given: Let coordinate of the point on the parabola be (x,y). Then,

y= x^2+7x +2                    ......(1)

Solution:  Let the distance of the point (x,(x^2+7x+2)) from the line y=3x-3 be S.

\begin{aligned} &\mathrm{S}=\left|\frac{-3 x+\left(x^{2}+7 x+2\right)+3}{\sqrt{10}}\right| \\ &\frac{d S}{d t}=\frac{-3+2 x+7}{\sqrt{10}} \\ &\frac{d S}{d t}=0 \\ &\frac{-3+2 x+7}{\sqrt{10}}=0 \\ &2 x=-4, x=-2 \\ &\frac{d^{2} S}{d t^{2}}=\frac{2}{\sqrt{10}}>0 \end{aligned}

The required nearest point is (x,(x^2+7x+2)) 




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