Question

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

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

Answer 1

Answer: (A)

General point on the curve $y^2=x-1$ is $(t_1^2+1,t_1)$ and the general point on the curve $x^2=y-1$ is $(t_2,t_2^2+1).$ Since both the curves are symmetrical about the line $y=x$ , for the nearest point on the curve $y^2=x-1$ from the line $y=x$
Let, $Q\left(t_1^2+1,t_1\right)$
$P(t_2,t_2^2+1)$
slopes of the tangents are same
$\frac{1}{2t_1}=2\cdot t_2$
$t_1{t}_2=\frac{1}{4}$ ......(1)
Since $PQ$ is perpendicular to line $x=y$
$\frac{t_1-t_2^2-1}{t_1^2+1-t_2}=-1$ ......(2)
Solving (1) & (2), we get $t_1=t_2=\frac{1}{2}$
$P\left(\frac{1}{2},\frac{5}{4}\right),Q\left(\frac{5}{4},\frac{1}{2}\right)$
Therefore, $PQ=\frac{3\sqrt{2}}{4}$ units