Question

Minimum area of circle which touches the parabolas $y=x^2+1$ and $y^2=x-1$ is

• A. $\frac{9\pi }{16}$ squnits
• B. $\frac{9\pi }{32}$ sq units
• C. $\frac{9\pi }{8}$ squnits
• D. $\frac{9\pi }{4}$ squnits

Answer 1

The correct option is B $\frac{9\pi }{32}$ sq units

Let the required circle touches parabola $y=x^2+1$ and $y=x^2-1$ at $A$ and $B$, respectively.
Here, both parabolas are symmetric w.r.t. $y=x$
Tangents at $A$ and $B$ will be parallel to $y=x$

For $y=x^2+1$
Slope of tangent $=2x_1=1$
$\Rightarrow x_1=\frac{1}{2}$
$\Rightarrow y_1=\frac{5}{4}$
So, $A\equiv (\frac{1}{2},\frac{5}{4})$

For $y^2=x-1$
Slope of tangent $=\frac{1}{2y_1}=1$
$\Rightarrow y_1=\frac{1}{2}$
$\Rightarrow x_1=\frac{5}{4}$
So, $B\equiv (\frac{5}{4},\frac{1}{2})$
$\Rightarrow AB=\frac{3\sqrt{2}}{4}$
Radius $=\frac{1}{2}AB=\frac{3\sqrt{2}}{8}$units
Area $=\frac{9\pi }{32}$ sq. units