Question

The shortest distance between the line $y=x-5$ and the parabola $y=x^2+3x+6$ is

• A. $\sqrt{2}\text{units}$
• B. $2\sqrt{2}\text{units}$
• C. $3\sqrt{2}\text{units}$
• D. $5\sqrt{2}\text{units}$

Answer 1

The correct option is D $5\sqrt{2}\text{units}$

Let $P(x_1,y_1)$ be the point on the parabola that is closest to the line $y=x-5$
Then $\frac{dy}{dx}{|}_{(x_1,y_1)}=$slope of the given line
$\Rightarrow 2x_1+3=1$
$x_1=-1\Rightarrow y_1=4$
Hence, the point $(-1,4)$ is closest and its perpendicular distance from the line $y=x-5$ will be the shortest distance. Therefore, shortest distance $=\frac{10}{\sqrt{2}}=5\sqrt{2}\text{units}$