Question

Shortest distance (in units) between the two parabolas $y^2=x-2,x^2=y-2$ is

• A. $\frac{1}{4\sqrt{2}}$
• B. $\frac{5}{4\sqrt{2}}$
• C. $\frac{7}{2\sqrt{2}}$
• D. $\frac{6}{7\sqrt{2}}$

Answer 1

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

The equations of the parabola are
$y^2=x-2\cdots \left(1\right)$
$x^2=y-2\cdots \left(2\right)$
We observe that if we interchange $x$ and $y$ in equation $\left(1\right),$ we obtain equation $\left(2\right)$. So, two parabolas are
symmetric about $y=x$.
Shortest distance exist between the tangents on both the parabolas which are parallel to $y=x$

Let $(x_1,y_1)$ and $(x_2,y_2)$ be the points on parabola $1$ and parabola $2$ respectively, from which they have the shortest distance between them.

For curve $\left(1\right)$, the equation of tangent is
$y{y}_1=\frac{x+x_1}{2}-2$
$\Rightarrow 2y{y}_1=x+x_1-4$
Slope $=\frac{1}{2y_1}=1$
$\Rightarrow y_1=\frac{1}{2}$

Putting value of $y_1$ in equation $\left(1\right)$, we get $x_1=\frac{9}{4}$
So, $(x_1,y_1)\equiv (\frac{9}{4},\frac{1}{2})$

Since, the parabolas are symmetric about $y=x,$
and if $(x_1,y_1)\equiv (\frac{9}{4},\frac{1}{2})$, then $(x_2,y_2)\equiv (\frac{1}{2},\frac{9}{4})$
$(\because \text{x and y will interchange})$
$\therefore d=\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2}$
$\Rightarrow d=\frac{7}{2\sqrt{2}}$