Question

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

• A. $(-2,-8)$
• B. $(1,10)$
• C. $(2,20)$
• D. $(-1,-4)$

Answer 1

Answer: (A)

$(-2,-8)$

Any point on the parabola is $(x,x^2+7x+2)$

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

$P=\left|\frac{3x-(x^2+7x+2)-3}{\sqrt{9+1}}\right|=\left|\frac{x^2+4x+5}{\sqrt{10}}\right|=\frac{x^2+4x+5}{\sqrt{10}}$

$($ as $x^2+4x+5>0$ for all $x\in R)$

$\frac{dP}{dx}=0\Rightarrow x=-2$.

So, the required point is $(-2,-8)$.