Question

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

• A. $2\sqrt{2}$
• B. $\frac{3\sqrt{2}}{4}$
• C. $4$
• D. $\sqrt{\frac{36}{5}}$

Answer 1

Answer: (B)

$\frac{3\sqrt{2}}{4}$

$y^2=x-1...............\left(1\right)$

$x^2=y-1................\left(2\right)$

$\left(1\right)$ and $\left(2\right)$ are inverse functions of each other and mirror images with respect to $x=y$

So, common normal both of them is always perpendicular to $y=x$. Slope of normal $=-1$

So, finding slopes of tangents on the foot of normal $=1$,

For $\left(1\right)$, $2y\frac{dy}{dx}=1$

$=>\frac{dy}{dx}=\frac{1}{2y}=-1\left(\frac{1}{-1}\right)=1$

$=>y=\frac{1}{2}$ and $x=\frac{5}{4}$

$A:(\frac{5}{4},\frac{1}{2})$

For, $\left(2\right)$, $2x=\frac{dy}{dx}=1$

$=>x=\frac{1}{2}$ and $y=\frac{5}{4}$

$AB=\sqrt{{(\frac{5}{4}-\frac{1}{2})}^2+({\frac{1}{2}-\frac{5}{4})}^2}$

$=\sqrt{2{\left(\frac{3}{4}\right)}^2}$

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