Question

If $x-1=0$ is the directix of parabola $y^2-kx+8=0$, then $k$ is equal to

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

Answer 1

Answer: (C)

$4$

We have, $y^2-kx-8=0$
$\Rightarrow y^2=k(x-\frac{8}{k})$
$\therefore$ Directrix $x-\frac{8}{k}=-\frac{k}{4}$
$\Rightarrow x-(\frac{8}{k}-\frac{k}{4})=0$
But equation of directrix is $x-1=0$. Hence, by comparing, we get
$\frac{8}{k}-\frac{k}{4}=1\Rightarrow k^2+4k-32=0$
$\Rightarrow (k-4)(k+8)=0$
$\Rightarrow k=4,-8$.