Question

Minimum distance between the curves $y^2=x-1$ and $x^2=y-1$ is equal to

• A. $\frac{3\sqrt{2}}{4}$
• B. $\frac{5\sqrt{2}}{4}$
• C. $\frac{7\sqrt{2}}{4}$
• D. $\frac{\sqrt{2}}{4}$

Answer 1

The correct option is A $\frac{3\sqrt{2}}{4}$

Clearly, the two curves are parabolas and symmetrical about the line $y=x.$, the distance of either of the parabolas from the line is equal and is half of the minimum distance between them.

Let $(x_1,y_1)$ and $(x_2,y_2)$ are the points on the the curves $y^2=x-1$ and $x^2=y-1$ respectively having minimum distance. Then at $(x_1,y_1)$ and $(x_2,y_2)$, slope is $1.$

$\therefore \frac{dy}{dx}=\frac{1}{2y_1}=1\Rightarrow y_1=\frac{1}{2},x_1=\frac{5}{4}$

and similarly for curve $2$,

$\frac{dy}{dx}=2x=1\Rightarrow x_2=\frac{1}{2},y_2=\frac{5}{4}$

$\therefore$ minimum distance $=\frac{3\sqrt{2}}{4}$