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

Option: 1

(-2,4)
 


Option: 2

(3,5)
 


Option: 3

(-2,-8)
 


Option: 4

(0,1)


Answers (1)

best_answer

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

Its distance from the line y=3 x-3 is given by 

            \begin{aligned} 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| \\ \\& =\frac{x^2+4 x+5}{\sqrt{10}}\left(\text { as } x^2+4 x+5>0, \text { for all } x \in R\right) \end{aligned}

   \begin{array}{ll} \therefore & \frac{d P}{d x}=0 \\ \\\Rightarrow & x=-2 \end{array}

\therefore The required point \equiv(-2,-8)

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rishi.raj

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