NIMCET 2014 Question Paper with solutions

We have
${\mathit{y}}^2=\mathit{k}\mathit{x}-8$
⇒ ${\mathit{y}}^2$ = k (x - $\frac{8}{\mathit{k}}$ )
which is a form of ${\mathit{y}}^2$ = 4AX
So, X = x - 8/k , A = k/4
∴ Equation of directrix , X = - A
⇒ x - $\frac{8}{\mathit{k}}$ = - $\frac{\mathit{k}}{4}$
⇒ x = $\frac{8}{\mathit{k}}-\frac{\mathit{k}}{4}$
⇒ 1 = $\frac{8}{\mathit{k}}-\frac{\mathit{k}}{4}$
⇒ 4k = 32 - ${\mathit{k}}^2$
⇒ ${\mathit{k}}^2$ + 4k - 32 = 0
⇒ ${\mathit{k}}^2$ + 8k - 4k -32 = 0
⇒ (k + 8) (k - 4) = 0
k = 4 , - 8