Question

If y=m1x+c and y=m2x+c are two tangents to the parabola y^2+4a(x+a)=0, then

Answer 1

Final Answer : $c = ma-\frac{1}{m}$

Steps:
1) We have,
Parabola
$y^2 +4a(x+a) =0$
where $y =m_1x+c \\ \\ y = m_2x+c$
are tangents .

Then,
This equation has real and equal roots.
$(mx+c)^2 + 4a(x+a) = 0 \\ \\ m^2x^2 +(2mc+4a)x + 4a^2+c^2 = 0 \\ \\ D = 0 \\ \\ =>16a^2+16mac-16m^2a^2 =0 \\ \\ =>a^2 +mca-m^2a^2=0 \\ \\ => c = ma-\frac{a}{m}$
where
$m = m_1 \quad or \quad m_2$
Hence, we got relation between c and m .