Question

The shortest distance between the line$y-x=1$ and the $x=y^2$ is

• A. $\frac{3\sqrt{2}}{8}$
• B. $\frac{2\sqrt{3}}{8}$
• C. $\frac{3\sqrt{2}}{5}$
• D. $\frac{\sqrt{3}}{4}$

Answer 1

The correct option is A

$\frac{3\sqrt{2}}{8}$

Explanation for the correct option.

Find the shortest distance.

Let the point on the curve be $\left({\alpha }^2,\alpha \right)$.

Then, the distance will be $d=\left|\frac{(\alpha -{\alpha }^2-1)}{\sqrt{2}}\right|$

$\Rightarrow d=\left|\frac{{\alpha }^2-\alpha +1}{\sqrt{2}}\right|....\left(1\right)$

Now, applying the concept of maxima and minima to find coordinates of the point.

$$\begin{gathered} \Rightarrow \frac{d\left(d\right)}{d\alpha }=\frac{d({\alpha }^2-\alpha +1)}{d\alpha } \\ \Rightarrow \frac{d\left(d\right)}{d\alpha }=2\alpha -1=0 \\ \Rightarrow \alpha =\frac{1}{2} \end{gathered}$$

Putting the value of $\alpha$ in $\left(1\right)$

$d=\left|\frac{{\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{2}}+1}{\sqrt{2}}\right|=\frac{3\sqrt{2}}{8}units$

Hence, option $A$ is the correct answer.