Question

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

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

Answer 1

The correct option is A $\frac{8}{3\sqrt{2}}$

$y^2=x\Rightarrow 2y=\frac{dx}{dy}\left[\text{Differentiating both sides w.r.t. 'y'}\right]{[-\frac{dx}{dy}]}_{x_1,y_1}=-2y_1$

$\therefore$ Slope of the normal of the parabola $y^x$ at $P(x_1,y_1)=-2y_1$

Again slope of the line y - x = 1 is 1.

$\therefore (-2y_1).\left(1\right)=-1\Rightarrow y_1=\frac{1}{2}$

Again $P(x_1,y_1)$ lie on $y^2=x$

$\therefore x_1=y_1^2=\frac{1}{4}.$

$\therefore$ The point P is $(\frac{1}{4},\frac{1}{2})$

$\therefore$ The distance from $(\frac{1}{4},\frac{1}{2})$ to y - x = 1

$d=\frac{|\frac{1}{2}-\frac{1}{4}-1|}{\sqrt{(1{)}^2+(-1{)}^2}}=\frac{|-\frac{1}{2}-\frac{1}{4}|}{\sqrt{2}}=\frac{3}{4\sqrt{2}}=\frac{3\sqrt{2}}{8}$