Question

A line $x-\frac{y}{\sqrt{3}}=c$ is drawn through the focus $\left(F\right)$ of the parabola $y^2-8x-16=0.$ If the two intersection points of the given line and the parabola are $P$ and $Q$, such that the perpendicular bisector of $PQ$ intersects the $x$-axis at $A$, then the length of $AF$ is

• A. $8$
• B. $\frac{8}{3}$
• C. $\frac{16}{3}$
• D. $\frac{8}{7}$

Answer 1

The correct option is C $\frac{16}{3}$


$y^2=8(x+2)\cdots \left(1\right)$
$\Rightarrow a=2$
Therefore, coordinates of $F$ is $(0,0)$
$x-\frac{y}{\sqrt{3}}=c$ passes through $(0,0)$
$\Rightarrow c=0\Rightarrow x-\frac{y}{\sqrt{3}}=0\Rightarrow \sqrt{3}x=y$
$\Rightarrow 3x^2=8(x+2)\left[\text{From (1)}\right]\Rightarrow 3x^2-8x-16=0\Rightarrow x=4,\frac{-4}{3}\Rightarrow y=4\sqrt{3},\frac{-4}{\sqrt{3}}$

Therefore, coordinates of intersection points are $P(4,4\sqrt{3}),Q(\frac{-4}{3},\frac{-4}{\sqrt{3}})$
Midpoint of $PQ$ is $(\frac{4}{3},\frac{4}{\sqrt{3}})$
Slope of $PQ$ is $\sqrt{3}$
$\therefore$ Slope of the perpendicular bisector is $\frac{-1}{\sqrt{3}}$

Equation of bisector :
$(y-\frac{4}{\sqrt{3}})=\frac{-1}{\sqrt{3}}(x-\frac{4}{3})$
For $A:y=0\Rightarrow x=\frac{16}{3}$

$A(\frac{16}{3},0)$
$\Rightarrow AF=\frac{16}{3}\text{units}$