Question

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

Given line is $y-x=1$

The shortest distance is the perpendicular distance from the line $y-x=1$ to the tangent of the curve parallel to the given line.

The general form of a line is $y=mx+c$ where $m$ is the slope of the line.

The line $y-x=1$ can be written as $y=x+1$

By comparing the equation of the line with general form we can say that slope of line $y-x=1$is $1$

We know that, slope of parallel lines are equal; so the slope of the line parallel to $y-x=1$ drawn to the curve is $1$

Now, to find the slope of the tangent to curve, differentiate $x=y^2$ with respect to $x$ as the slope of the tangent is given by $\frac{dy}{dx}$.

$\therefore \frac{d}{dx}\left(x\right)=\frac{d}{dx}\left(y^2\right)$

$$\begin{gathered} \Rightarrow 1=2y\frac{dy}{dx} \\ \Rightarrow y\frac{dy}{dx}=\frac{1}{2} \end{gathered}$$

Since the slope of given parallel lines$=1$ i.e. slope of the required tangent $\frac{dy}{dx}=1$

$$\begin{gathered} \therefore y\left(1\right)=\frac{1}{2} \\ \Rightarrow y=\frac{1}{2} \end{gathered}$$

On substituting the value of $y$ in equation of curve, we get,

$x=y^2=\frac{1}{4}$

Therefore,$\left(\frac{1}{4},\frac{1}{2}\right)$ is a point on the curve through which a tangent parallel to the given line $y-x=1$ passes

We know that the shortest distance between the point $\left(x_1,y_1\right)$ and the line $ax+by+c=0$ is given by $\frac{\left|a{x}_1+b{y}_1+c\right|}{\sqrt{a^2+b^2}}$

Therefore, shortest distance of the point $\left(\frac{1}{4},\frac{1}{2}\right)$ from $y-x=1$is,

$\begin{array}{ccc}\frac{\left|\left({\displaystyle \frac{1}{2}}\right)-\left({\displaystyle \frac{1}{4}}\right)-1\right|}{\sqrt{1+1}}& =& \frac{\left|{\displaystyle \frac{2-1-4}{4}}\right|}{\sqrt{2}}\\ & =& \frac{3}{4\sqrt{2}}\\ & =& \frac{3}{4\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}\\ & =& \frac{3\sqrt{2}}{8}\end{array}\begin{array}{}\\ \\ \\ \end{array}$

Thus, the shortest distance is $\frac{3\sqrt{2}}{8}$.

Hence, the correct option is option(A) i.e. $\frac{3\sqrt{2}}{8}$