Question

Minimum distance between the curves $y^2=x-1$ and $x^2=y-1$ is $\frac{a\sqrt{2}}{b}$, where $a,b$ are co-prime numbers, then the value of $a+b$ is

Answer 1

Given : $y^2=x-1$ and $x^2=y-1$
By observation, if we swap $x$ by $y$ in $y^2=x-1$, then we get $x^2=y-1$
So, they are symmetric about $y=x$
Finding the equation of tangents on both the curve which is parallel to $y=x$
Now for $y^2=x-1$,
$2y{y}^{\prime }=1\Rightarrow y^{\prime }=\frac{1}{2y}=1\Rightarrow y=\frac{1}{2}\Rightarrow x=\frac{5}{4}\Rightarrow x-y-\frac{3}{4}=0$
Now for $x^2=y-1$,
$y^{\prime }=2x\Rightarrow 2x=1\Rightarrow x=\frac{1}{2}\Rightarrow y=\frac{5}{4}\Rightarrow x-y+\frac{3}{4}=0$

Distance between parallel lines
$=\left|\frac{-\frac{3}{4}-\frac{3}{4}}{\sqrt{1^2+1^2}}\right|=\frac{3\sqrt{2}}{4}$

Hence, $a+b=7$