Question

As is shown in figure, $P\left(x_0,y_0\right)$ is a moving point on the parabola $y^2=2x$,point $B$ and $C$ are on $y$-axis, and the sides of triangle PBC touches the $\mathrm{circle}(x-1{)}^2+y^2=1$.

The minimum value of area of triangle $PBC$ is___ (in sq. unit)

Answer 1

Answer: 8

Let point $B(0,b)$ and $C(0,c)$ and assume $b>c$
The equation of $PB$ is $y-b=\frac{y_0-b}{x_0}(x)$
$\Rightarrow \left(y_0-b\right)x-x_{0}y+x_{0}b=0$ as its tangent of circle $(x-1{)}^2+y^2=1$
$\Rightarrow 1=\left|\frac{y_0-b+x_{0}b}{\sqrt{{\left(y_0-b\right)}^2+x_0^2}}\right|\Rightarrow \left(x_0-2\right)b^2+2y_{0}b-x_0=0$ similarly $\left(x_0-2\right)c^2+2y_{0}c-x_0=0$
$\Rightarrow b+c=\frac{-2y_0}{x_0-2}$ and $bc=\frac{-x_0}{x_0-2}\Rightarrow (b-c{)}^2=\frac{4x_0^2+4y_0^2-8x_0}{{\left(x_0-2\right)}^2}\Rightarrow b-c=\frac{2x_0}{x_0-2}$ (as $y_0^2=2x_0)$
Now, area of $△PBC=\frac{1}{2}(b-c)x_0=\frac{x_0^2}{x_0-2}=\left(x_0-2\right)+\frac{4}{\left(x_0-2\right)}+4\ge 8$