Parabola-3

Posted February 2nd, 2010 by alpha

If y=mx+c touches the parabola $y^2=4a(x+a)$ , then c=?

Comments

On February 4th, 2010 Structure says:

Equation

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

has to have double real solutions.

$m^2x^2+2mcx-4ax+c^2-4a^2=0$

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

$4amc=4a^2+4a^2m^2$

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