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Q.
The straight line $y=m x+c(m>0)$ touches the parabola $y^{2}=8(x+2)$. Then the minimum value taken by $c$ is
Conic Sections
Solution:
Tangent to $y^{2}=8(x+2)$ having slope ' $m$ ' is
$y=m(x+2)+\frac{2}{m} $
$\therefore c=2 m+\frac{2}{m}$
$\Rightarrow \frac{c}{2}=\left(m+\frac{1}{m}\right) $
$\because m+\frac{1}{m} \geq 2 $
$\Rightarrow \frac{c}{2} \geq 2 \Rightarrow c \geq 4$
Thus, the minimum value of $c$ is $4$ .