The line y−3x+3=0 cuts the parabola y2=x+2 at A and B. If P=(3,0) then PA.PB is equal to

A point on the line is x=r2+3,y=32rPutting these values of x and y in the equation of parabola we get32r2=r2+3+2⇒3r2−2r−4(3+2)=0If r1 and r2 are its roots, then PA= r1PB=r2 and r1r2=43(3+2)Since the point P lies inside the parabole. So r1 and r2 are the opposite signsSo r1r2<0. Then r1r2=43(3+2), Hence PA.PB =43(3+2)

The line y−3x+3=0 cuts the parabola y2=x+2 at A and B. If P=(3,0) then PA.PB is equal to

A point on the line is x=r2+3,y=32rPutting these values of x and y in the equation of parabola we get32r2=r2+3+2⇒3r2−2r−4(3+2)=0If r1 and r2 are its roots, then PA= r1PB=r2 and r1r2=43(3+2)Since the point P lies inside the parabole. So r1 and r2 are the opposite signsSo r1r2<0. Then r1r2=43(3+2), Hence PA.PB =43(3+2)

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

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$(\frac{2}{3}) (2 + \sqrt{3})$

$(\frac{4}{3}) (2 - \sqrt{3})$

$(\frac{4}{3}) (2 + \sqrt{3})$

$2 \sqrt{3}$

answer is A.

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Detailed Solution

A point on the line is $x = \frac{r}{2} + \sqrt{3}, y = \frac{\sqrt{3}}{2} r$
Putting these values of x and y in the equation of parabola we get
${(\frac{\sqrt{3}}{2} r)}^{2} = \frac{r}{2} + \sqrt{3} + 2 \Rightarrow 3 r^{2} - 2 r - 4 (\sqrt{3} + 2) = 0$
If r₁ and r₂ are its roots, then PA= r₁
$PB = r_{2} and r_{1} r_{2} = \frac{4}{3} (\sqrt{3} + 2)$
Since the point P lies inside the parabole. So r₁ and r₂ are the opposite signs
So $r_{1} r_{2} < 0 . Then r_{1} r_{2} = \frac{4}{3} (\sqrt{3} + 2), Hence PA.PB = \frac{4}{3} (\sqrt{3} + 2)$