Question

If the line $y-2=0$ is the directrix of the parabola $x^2-ky+32=0,k\ne 0$ 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

Answer: 16

$x^2-ky+32=0$
$\Rightarrow x^2=k\left(y-\frac{32}{k}\right)$
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 $k=\left|-16\right|=16$