Prove that the straight line y = mx + c touches the parabola $y^{2} = 4 a (x + a)$ if $c = m a + \frac{a}{m} .$

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Solution

satisfy the given line with parabola
and then put $D = b^{2} - 4 a c = 0$ quad eqn. $a x^{2} + b x + c$
parabola
$y^{2} = 4 a (x + a)$ line $y = m x + c$
$(m x + c)^{2} = 4 a (x + a)$
$(m x)^{2} + 2 m x c + c^{2} = 4 a (x + a)$
$(m x)^{2} + 2 m x c + c^{2} - 4 a x - 4 a^{2} = 0$
Now put D = 0
$D = b^{2} - 4 a c = 0$
$(2 m c - 4 a)^{2} - 4 m^{2} (c^{2} - 4 a^{2}) = 0$
$a^{2} - m c a + m^{2} a^{2} = 0$
$a^{2} + m^{2} a^{2} = m c a$
$c = \frac{a}{m} + m a$

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