Prove that the straight line Ix + my + n = 0 touches the parabola $y^{2} = 4 a x$ if $l n = a m^{2} .$

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Solution

straight line $l x + m y + n = 0$ is tangent to
parabola $y^{2} = 4 a x$

if line $y = M x + c$ touches parabola $y^{2} = 4 a x$
then
$c = \frac{a}{M}$
for the given line $\frac{- l x - n}{m} = y$
Now

$\frac{- n}{m} = \frac{a}{\frac{- l}{m}}$
$\Rightarrow l n = a m^{2}$

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l x + m y + n = 0

y^{2} = 4 a x

l n = a m^{2}

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y^{2} = 4 a x

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l x + m y + n = 0

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