Question

Shortest distance between the curves $y^2=x-2$ and$y=x$ is

• A. greater than $4$
• B. less than $2$
• C. greater than $3$
• D. greater than $2$

Answer 1

The correct option is B

less than $2$

Explanation for the correct answer:

Finding the shortest distance between the given curves:

We have $y^2=x-2$

Differentiating the above equation we get,

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

Slope of the line $y=x$ is $1$.

The slope of the tangent to the curve $=\frac{1}{2}y$

$$\begin{gathered} \Rightarrow \frac{1}{2y}=1 \\ \Rightarrow y=\frac{1}{2} \\ x=\frac{1}{4}+2\mathbf{[}\mathbf{\because }{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{]} \end{gathered}$$

$\therefore N=(\frac{9}{4},\frac{1}{2})$

Minimum distance $=MN=$ the perpendicular distance from the point $N$ to line $y=x$, given as:

$$\begin{gathered} \Rightarrow \left|\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}\right| \\ \Rightarrow \left|\frac{1\left({\displaystyle \frac{9}{4}}\right)-1\left({\displaystyle \frac{1}{2}}\right)}{\sqrt{1+1}}\right| \\ \Rightarrow \left|\frac{{\displaystyle \frac{9-2}{4}}}{\sqrt{2}}\right| \\ \Rightarrow \left|\frac{7}{4\sqrt{2}}\right| \end{gathered}$$

which is less than $2$.

Therefore, the correct answer is option (B).