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