If $m$ varies then find the range of $c$ for which the line $y=mx + c$ touches the parabola $y^2 = 8(x+2)$ .
My Attempt:
Put the value $y = mx + c$ in the parabola equation and then done $\Delta = 0$ or $\Delta >0$
I am getting $16/(m^2 + 8m) >0$
But in this, how do I neglect $m$?
The problem is asking you to find for every $m$ the range of $c$ such that the line touches the parabola. Substituting $x=\frac{y-c}{m}$ in the parabola equation we get $$my^2-8y-16m+8c=0$$ so $$\Delta=64+64m^2-32mc=32(2m^2-mc+2).$$ You need $\Delta \ge0$, hence $2m^2-mc+2\ge0$, that is equivalent to: $$c\le 2m+\frac{2}{m}$$ when $m>0$ $$c\ge 2m+\frac{2}{m}$$ when $m<0$.