Question

If the line $x-1=0$ is the directrix of the parabola $y^2-kx+8=0,$ then $k$ can be

• A. $\frac{1}{8}$
• B. $-8$
• C. $4$
• D. $\frac{1}{4}$

Answer 1

Answer: (B), (C)

$4$

The given parabola is
$y^2=kx-8=k(x-\frac{8}{k})$
Above parabola is in the form of $Y^2=kX$ where $X=x-\frac{8}{k}$ and $Y=y$.
Directrix of this parabola is $X=-\frac{k}{4}$
i.e., $x=-\frac{k}{4}+\frac{8}{k}$
This will coincide with $x=1$ if
$\frac{8}{k}-\frac{k}{4}=1\Rightarrow 32-k^2=4k\Rightarrow k^2+4k-32=0$
$\Rightarrow (k+8)(k-4)=0$
$\Rightarrow k=-8,4$