Question

If the line $y-\sqrt{3}x+3=0$ cuts the parabola $y^2=x+2$ at A and B, then $PA.PB$ is equal to [where $P\equiv (\sqrt{3},0)$]

• A. $\frac{4(\sqrt{3}+2)}{3}$
• B. $\frac{4(2-\sqrt{3})}{3}$
• C. $\frac{4\left(\sqrt{3}\right)}{2}$
• D. $\frac{2(\sqrt{3}+2)}{3}$

Answer 1

The correct option is A $\frac{4(\sqrt{3}+2)}{3}$

Consider the equation of the line.

$y=\sqrt{3}x-3$

Consider the equation of the parabola.

$y^2=x+2$

And,

$\Rightarrow P(\sqrt{3},0)$

Find the intersection points.

${(\sqrt{3}x-3)}^2=x+2$

$3x^2+9-6\sqrt{3}x=x+2$

$3x^2-x(6\sqrt{3}+1)+7=0$

$\Rightarrow x_1+x_2=\frac{6\sqrt{3}+1}{3}......\left(1\right)$

$\Rightarrow x_1{x}_2=\frac{7}{3}......\left(2\right)$

Any point the parabola can be written as,

$\Rightarrow (x,\sqrt{x+2})$

Let,

$A=(x_1,\sqrt{x_1+2})$

$B=(x_2,\sqrt{x_2+2})$

Therefore,

$PA=\sqrt{{(\sqrt{3}-x_1)}^2+{\left(\sqrt{x_1+2}\right)}^2}$

$\Rightarrow PA=\sqrt{3+x_1^2-2\sqrt{3}{x}_1+x_1+2}$

$PB=\sqrt{{(\sqrt{3}-x_2)}^2+{\left(\sqrt{x_2+2}\right)}^2}$

$\Rightarrow PB=\sqrt{3+x_2^2-2\sqrt{3}{x}_2+x_2+2}$

Therefore,

$PA\times PB=\sqrt{(5+x_1^2-(2\sqrt{3}-1)x_1)(5+x_2^2-(2\sqrt{3}-1)x_2)}$

$PA\times PB=\frac{4}{3}(\sqrt{3}+2)=\frac{4(\sqrt{3}+2)}{3}$

Hence, this is the required result.