I slipped off the parabola!

First, we can prove the first result. $y=mx+c$ will be a tangent to the parabola $y^2=4ax$ if and only if the system with these two equation has only one solution. So let's solve it:

$(mx+c)^2=4ax$

$m^2x^2+(2mc-4a)x+c^2=0$

which has only one solution if and only if $\frac{\Delta} {4}=0$ Therefore:

$\frac{\Delta} {4}=m^2c^2+4a^2-4amc-m^2c^2=0$ $\Rightarrow c=\frac{a} {m}$

Similarly we can obtain the second result:

$(mx+c)^2=4ax+4a^2$

$m^2x^2+(2mc-4a)x+(c^2-4a^2)=0$

$\frac{\Delta} {4}=m^2c^2+4a^2-4amc-m^2c^2+4a^2m^2=0$

$\Rightarrow c=\frac{a(1+m^2)} {m}=a(m+\frac{1} {m})$