Question

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

• A. $2\sqrt{2}$
• B. $\frac{7}{2\sqrt{2}}$
• C. $4$
• D. $\sqrt{\frac{36}{5}}$

Answer 1

The correct option is B $\frac{7}{2\sqrt{2}}$

Minimum distance is along the common normal.

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

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

$\left(1\right)$ and $\left(2\right)$ are inverse functions of each other so they mirror image of each other along $x=y.$ So, common normals is also perpendicular to $x=y.$ So, slope of normal will be $=-1$

Slope of tangent to $\left(1\right)=>\frac{d}{y}dx=\frac{1}{2}y=\frac{-1}{-1}=1$

$=>y=\frac{1}{2}$

and $y=\frac{9}{4}$

Point on $\left(1\right)$ $A(\frac{9}{4},\frac{1}{2})$ and on $\left(2\right)$ $B$ $(\frac{1}{2},\frac{9}{4})$

$AB=\sqrt{({\frac{9}{4}-\frac{1}{2})}^2+{(\frac{1}{2}-\frac{9}{2})}^2}=\sqrt{2({\frac{9}{4}-\frac{1}{2})}^2}$

$=\sqrt{\frac{49}{8}}$

$=\frac{7}{2\sqrt{2}.}$