Question

If the line $y-2=0$ is the directrix of the parabola $x^{2}-ky+32=0,k\neq 0$ and the parabola intersects the circle $x^{2}+y^{2}=8$ at two real distinct points, then the absolute value of $k$ is

Answer 1

$x^{2}-ky+32=0$
$\Rightarrow x^{2}=k\left(y - \frac{32}{k}\right)$
Put, $x=X,y-\frac{32}{k}=Y$
The equation of directrix is $Y+\frac{k}{4}=0$
i.e. $y-\frac{32}{k}+\frac{k}{4}=0$
But, $y-2=0$ is the directrix.
$\Rightarrow \frac{32}{k}-\frac{k}{4}=2$
$\Rightarrow k^{2}+8k-128=0$
$\Rightarrow k=-16$ or $k=8$
For $k=8,$ the parabola is $x^{2}=8\left(y - 4\right)$ which does not intersect the circle.
For $k=-16,$ the parabola is $x^{2}=-16\left(y + 2\right)$ which intersects the circle at two real distinct points.
$\Rightarrow $ Absolute value of $k=\left|- 16\right|=16$