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The coordinates of the point on the parabola \mathrm{y=x^{2}+7 x+2}, which is nearest to the straight line \mathrm{y=3 x-3} are

Option: 1

(-2,-8)


Option: 2

(1,10)


Option: 3

(2,20)


Option: 4

(-1,-4)


Answers (1)

best_answer

Any point on the parabola is \mathrm{\left(x, x^{2}+7 x+2\right)}

Its distance from the line \mathrm{y=3 x-3} is given by

\mathrm{P=\left|\frac{3 x-\left(x^{2}+7 x+2\right)-3}{\sqrt{9+1}}\right|=\left|\frac{x^{2}+4 x+5}{\sqrt{10}}\right|}
\mathrm{=\frac{x^{2}+4 x+5}{\sqrt{10}}\left(\text { as } x^{2}+4 x+5>0 \text { for all } x \in R\right)}
\mathrm{\frac{d P}{d x}=0 \Rightarrow x=-2}

The required point \mathrm{\equiv(-2,-8)}

Hence (A) is the correct answer.

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