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

Answer: 41

$\Rightarrow$ The given parabola are symmetric about the line $y=x$
$\Rightarrow$ Tangents at $A$ and $B$ must be parallel to $y=x$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{minA}=1={\left(\frac{dy}{dx}\right)}_{minB}$
$\iff 2x_B=1\Rightarrow x_h=\frac{1}{2}\Rightarrow y_B=x_B^2+1=\frac{5}{4}$
$\Rightarrow B=\left(\frac{1}{2},\frac{5}{4}\right)\Rightarrow A=\left(\frac{5}{4},\frac{1}{2}\right)$
$\Rightarrow |AB|=\sqrt{{\left(\frac{5}{4}-\frac{1}{2}\right)}^2+{\left(\frac{1}{2}-\frac{5}{4}\right)}^2}$
$=\frac{3\sqrt{2}}{4}$
$\Rightarrow$ Radius of a circle $=\frac{3\sqrt{2}}{8}$
$\Rightarrow$ Required area $=\frac{\pi (9)(2)}{64}=\frac{9\pi }{32}$
$\Rightarrow a=9,b=32$
$\Rightarrow a+b=41$