Question

Minimum area of circle which touches the parabola's $y=x^2+1$ and $y^2=x-1$ is $A$ . Evaluate
$\frac{32}{\pi }A$

Answer 1

Given,

a circle of minimum possible area is enclosed between the two parabola's

$y=x^2+1$ and $y^2=x-1$

The parabolas $y=x^2+1$ and $y^2=x-1$ are symmetrical about the line $y-x$. So the circle with minimum area touching both the parabolas will lie between the tangents to the parabola which are parallel to the line $y=x$.

$\therefore$ the slope of tangent to the curve $y^2=x-1$ at a point $A$, such that the tangent at ${}^{\prime }{A}^{\prime }$ is parallel to the line $y=x$ is

$2y\frac{dy}{dx}=1$

$\Rightarrow \frac{dy}{dx}=\frac{1}{2y}$

Now, $\frac{dy}{dx}=1$ $($ $\because$ tangent is $\parallel$ to $y=x)$

$\Rightarrow \frac{1}{2y}=1$

$\Rightarrow y=1/2$

$\therefore$ $x=5/4$ $\therefore$ $A(\frac{5}{4},\frac{1}{2})$

Similarly, the slope of tangent to the curve $y=x^2+1$ at a point $B$, such that the tangent is parallel to $y=x$ is :

$\frac{dy}{dx}=1$

$\Rightarrow 2x=1$

$\Rightarrow x=1/2$ $\therefore$ $y=5/4$ $\therefore$ $B=(\frac{1}{2},\frac{5}{4})$

$\therefore$ Radius $=\frac{1}{2}\sqrt{{(\frac{1}{2}-\frac{5}{4})}^2+{(\frac{5}{4}-\frac{1}{2})}^2}=\frac{1}{2}\sqrt{\frac{9}{16}+\frac{9}{16}}=\frac{3}{8}\sqrt{2}$

$\therefore$ Area of the circle $=A=\Pi {\left(Radius\right)}^2=\Pi .\frac{9}{32}$

$\therefore$ The value of $\frac{32}{\Pi }A=\frac{32}{\Pi }\times \Pi \times \frac{9}{32}=9$