If the straight line $y=m x+c$ touches the parabola $y^2-4 a x+4 a^3=0$, then $c$ is
If the straight line $y=m x+c$ touches the parabola $y^2-4 a x+4 a^3=0$, then $c$ is
$a m+\frac{a}{m}$
$a m-\frac{a}{m}$
$\frac{a}{m}+a^2 m$
$\frac{a}{m}-a^2 m$
Solution
Solving the given equations,
$
\begin{aligned}
&(m x+c)^2=4 a x-4 a^3 \\
& \Rightarrow m^2 x^2+2 m c \cdot x+c^2=4 a x-4 a^3 \\
& \Rightarrow m^2 x^2+(2 m c-4 a) x+c^2+4 a^3=0
\end{aligned}
$
Since the straight line touches the parabola at a point, so, the discriminant $=0$
$
\begin{aligned}
& \Rightarrow \quad(2 m c-4 a)^2-4 m^2\left(c^2+4 a^3\right)=0 \\
& \Rightarrow 4 m^2 c^2-16 a m c+16 a^2-4 m^2 c^2-16 a^3 m^2=0 \\
& \Rightarrow-m c+a-a^2 m^2=0 \\
& \Rightarrow m c=a-a^2 m^2 \Rightarrow c=\frac{a}{m}-a^2 m
\end{aligned}
$