If $m$ varies then, let the range of $c$ for which the line $y = m x + c$ touches the parabola $y^{2} = 8 (x + 2)$ be $(- \infty, - k] \cup [k, \infty)$ . Find $\frac{k}{2}$

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Solution

$y = m x + c$ touches the parabola $y^{2} = 8 (x + 2)$
$∴ (m x + c)^{2} = 8 (x + 2)$
$m^{2} x^{2} + x (2 m c - 8) + c^{2} - 16 = 0$ ...... (i)
line touch the parabola so D = 0 of equation (i)
$4 (m c - 4)^{2} - 4 m^{2} (c^{2} - 16) = 0$
$m^{2} c^{2} - 8 m c + 16 - m^{2} c^{2} + 16 m^{2} = 0$
$2 m^{2} - m c + 2 = 0$
Since m is real $D \geq 0$
$c^{2} - 16 \geq 0$
$c \in (- \infty, - 4] \cup [4, \infty)$
$∴ \frac{k}{2} = 2$

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