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Q.
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
Conic Sections
Solution:
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 $- 8$
$\Rightarrow \, k=4$