Question

As is shown in figure, $P \left( x _0, y _0\right)$ is a moving point on the parabola $y ^2=2 x$,point $B$ and $C$ are on $y$-axis, and the sides of triangle PBC touches the $\operatorname{circle}(x-1)^2+y^2=1$.

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

Answer 1

Let point $B(0, b)$ and $C(0, c)$ and assume $b>c$
The equation of $P B$ 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 $(x-1)^2+y^2=1$
$\Rightarrow 1=\left|\frac{y_0-b+x_0 b}{\sqrt{\left(y_0-b\right)^2+x_0^2}}\right| \Rightarrow\left(x_0-2\right) b^2+2 y_0 b-x_0=0$ similarly $\left(x_0-2\right) c^2+2 y_0 c-x_0=0$
$\Rightarrow b + c =\frac{-2 y _0}{ x _0-2}$ and $bc =\frac{-x_0}{x_0-2} \Rightarrow(b-c)^2=\frac{4 x_0^2+4 y_0^2-8 x_0}{\left(x_0-2\right)^2} \Rightarrow b-c=\frac{2 x_0}{x_0-2}$ (as $\left.y_0^2=2 x_0\right)$
Now, area of $\triangle PBC =\frac{1}{2}( b - c ) x _0=\frac{x_0^2}{x_0-2}=\left(x_0-2\right)+\frac{4}{\left(x_0-2\right)}+4 \geq 8$