Question

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

• A. $1/8$
• B. $8$
• C. $4$
• D. $1/4$

Answer 1

Answer: (C)

The given parabola is $y^2=kx-8=k\left(x-\frac{8}{k}\right)$.
Shifting the origin to $\left(\frac{8}{k},0\right)$.
Equation of the parabola becomes $Y^2=4\frac{k}{4}X$,
where $X=x-\frac{8}{K}$ and $y=Y$.
Directrix of this parabola is $X=\frac{-k}{4}$ or $x-\frac{8}{k}=\frac{-k}{4}$
This will be coincident with $x=1$, if $\frac{8}{k}-\frac{k}{4}=1$
$\Rightarrow k^2+4k-32=0$
$\Rightarrow \left(k+8\right)\left(k-4\right)=0$
$\Rightarrow k=4$ or $-8$
$\Rightarrow k=4$