Question

Consider the parabola $y=x^2+7x+2$ and the straight line $y=3x-3$. What is the shortest distance from the above point on the parabola to the line?

• A. $\frac{\sqrt{10}}{2}$
• B. $\frac{\sqrt{10}}{5}$
• C. $\frac{1}{\sqrt{10}}$
• D. $\frac{\sqrt{5}}{4}$

Answer 1

Answer: (C)

$\frac{1}{\sqrt{10}}$

Let $(h,k)$ be the point closest to the line.

So, the tangent at $(h,k)$ must be parallel to the given line

$y=x^2+7x+2$

$\therefore \frac{dy}{dx}=2x+7$

Slope of line $y=3x-3$ is $3$

$\therefore 2h+7=3$

$\therefore h=-2$

$\therefore k=h^2+7h+2=4-14+2=-8$

Distance of point $(-2,-8)$ from the line $=\left|\frac{3(-2)-(-8)-3}{\sqrt{3^2+(-1{)}^2}}\right|=\frac{1}{\sqrt{1}0}$

So, the answer is option (C).