Question

The co-ordinate of a point on the parabola $y=x^2+7x+2$ which is nearest to the straight line $y=3x-3$ is

• A. $(2,8)$
• B. $(-2,8)$
• C. $(-2,-8)$
• D. $(2,-8)$

Answer 1

The correct option is C $(-2,-8)$

Let $P$ be the point on the line closest to the parabola and let the shortest distance line the ray $P$ meet the parabola in $Q$.
Then $PQ$, is the shortest distance between the line and the parabola.

$RQ$ must be perpendicular to the given line and must be normal to the parabola at $Q$.
Let the equation of the line $PQ$ be $=-\frac{1}{3}x+k$ (Since $PQ$ is $\perp$ to the given line)

Let the co-ordinates of $Q$ be $(x_1,y_1)$
Slope of the tangent to the parabola at $x=x_1$ is given by $\frac{dy}{dx}=2x_1+7$

$\Rightarrow$Slope of the normal at $x=x_1$ is $\frac{-1}{(2x_1+7)}$

$\Rightarrow \frac{-1}{2x_1+7}=-\frac{1}{3}\Rightarrow x_1=-2$
$\therefore y_1={x_1}^2+7x_1+2=4-14+2=-8$

$Q$ is $(-2,-8)$