Mathematics - Parabola Question with Solution | TestHub

The line

$y-\sqrt{3}x+3=0$

can be written as

$y=\sqrt{3}x-3$. The slope is

$m=\sqrt{3}$, so the angle

$\theta$ the line makes with the positive x-axis is

$\arctan (\sqrt{3})=60^{\circ }$. A point on the line at a directed distance

$r$ from

$P(\sqrt{3},0)$ has coordinates

$(x,y)$ given by the parametric equations:

$x=\sqrt{3}+r\cos (60^{\circ })=\sqrt{3}+\frac{r}{2}$

$y=0+r\sin (60^{\circ })=\frac{r\sqrt{3}}{2}$

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Step 2: Formulate the quadratic equation for distances

Substitute the parametric equations for

$x$ and $y$ into the parabola equation

$y^2=x+2$:

$(\frac{r\sqrt{3}}{2}{)}^2=(\sqrt{3}+\frac{r}{2})+2$

$\frac{3r^2}{4}=\sqrt{3}+\frac{r}{2}+2$

Multiply by 4 to clear the denominator:

$3r^2=4\sqrt{3}+2r+8$

Rearrange into a standard quadratic form

$A{r}^2+Br+C=0$:

$3r^2-2r-(8+4\sqrt{3})=0$

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Step 3: Calculate the product of distances

The roots

$r_1$ and $r_2$ of this quadratic equation are the directed distances from point

$P$ to points

$A$ and $B$. The actual distances are

$PA=|r_1|$ and $PB=|r_2|$. The product of the roots is given by the formula

$r_1{r}_2=\frac{C}{A}$:

$r_1{r}_2=\frac{-(8+4\sqrt{3})}{3}$

The product of the distances

$PA\cdot PB$ is the absolute value of the product of the roots:

$PA\cdot PB=|r_1{r}_2|=|\frac{-(8+4\sqrt{3})}{3}|=\frac{8+4\sqrt{3}}{3}$

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**Answer:** The product

$PA\cdot PB$ is equal to

$\frac{8+4\sqrt{3}}{3}$.