Question

If the minimum area of the circle which touches the parabola $y=x^{2}+1$ and $y^{2}=x-1$ is $\frac{a \pi }{b}$ sq.units , where $a\&b$ are co-prime numbers, then the value of $a+b$ is

Answer 1

$\Rightarrow $ The given parabola are symmetric about the line $y=x$
$\Rightarrow $ Tangents at $A$ and $B$ must be parallel to $y=x$
$\Rightarrow \left(\frac{d y}{d x}\right)_{min A}=1=\left(\frac{d y}{d x}\right)_{min B}$
$\Leftrightarrow 2x_{B}=1\Rightarrow x_{h}=\frac{1}{2}\Rightarrow y_{B}=x_{B}^{2}+1=\frac{5}{4}$
$\Rightarrow B=\left(\frac{1}{2} , \frac{5}{4}\right)\Rightarrow A=\left(\frac{5}{4} , \frac{1}{2}\right)$
$\Rightarrow \left|\right.AB\left|\right.=\sqrt{\left(\frac{5}{4} - \frac{1}{2}\right)^{2} + \left(\frac{1}{2} - \frac{5}{4}\right)^{2}}$
$=\frac{3 \sqrt{2}}{4}$
$\Rightarrow $ Radius of a circle $=\frac{3 \sqrt{2}}{8}$
$\Rightarrow $ Required area $=\frac{\pi \left(\right. 9 \left.\right) \left(\right. 2 \left.\right)}{64}=\frac{9 \pi }{32}$
$\Rightarrow a=9,b=32$
$\Rightarrow a+b=41$