Question

Shortest distance between the line $y-x=1$ and the curve $x=y^2$ is

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

Answer 1

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

Given equation of parabola is $x=y^2$

Let $P(h,k)$ be a point on the parabola

$\Rightarrow h=k^2$

Distance of the line $x-y+1=0$ from the point $P(h,k)$ is

$D=\left|\frac{h-k+1}{\sqrt{2}}\right|$

$\Rightarrow D=\left|\frac{k^2-k+1}{\sqrt{2}}\right|$

$\Rightarrow D=\left|\frac{{(k-\frac{1}{2})}^2+\frac{3}{4}}{\sqrt{2}}\right|$

Since, numerator and denominator both are positive,

So, $D=\frac{{(k-\frac{1}{2})}^2+\frac{3}{4}}{\sqrt{2}}$

For maxima or minima,

$\frac{dD}{dk}=0$

$\Rightarrow k-\frac{1}{2}=0\Rightarrow k=\frac{1}{2}$

Also,$\frac{d^{2}D}{d{k}^2}>0$

Hence, D is minimum at $k=\frac{1}{2}$

$\Rightarrow h=\frac{1}{4}$

Hence, the point on the curve is $(\frac{1}{4},\frac{1}{2})$

$D=\frac{3}{4\sqrt{2}}=\frac{3\sqrt{2}}{8}$