Question

A line $x+1=y$ meets the curve $2x^3+10x^2+x-4=y$ at $A,B$ and $C$. If point $P\equiv (-1,0)$, then $|PA.PB.PC|$ is equal to

• A. $3\sqrt{2}$
• B. $3$
• C. $3\sqrt{3}$
• D. $2\sqrt{3}$

Answer 1

The correct option is A $3\sqrt{2}$

Parametric form of the line $y=x+1$ is
$\frac{y}{\cos 45°}=\frac{x+1}{\sin 45°}=r\Rightarrow x=\frac{r}{\sqrt{2}}-1,y=\frac{r}{\sqrt{2}}$
Putting the values of $x,y$ in the equation of the curve
we get $2{(\frac{r}{\sqrt{2}}-1)}^3+10{(\frac{r}{\sqrt{2}}-1)}^2+(\frac{r}{\sqrt{2}}-1)-4=\left(\frac{r}{\sqrt{2}}\right)\Rightarrow \frac{r^3}{\sqrt{2}}+2r^2-7\sqrt{2}r+3=\mathrm{0...}\left(1\right)$
If the roots of the equation are $r_1,r_2\&r_3\therefore r_1{r}_2{r}_3=-3\sqrt{2}\therefore |PA.PB.PC|=3\sqrt{2}$