The condition that the straight line $l x + m y + n = 0$ touches the parabola $x^{2} = 4 a y$ is

b n = a m^{2}

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b l^{2} - m n = 0

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l n = a m^{2}

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a m = l n^{2}

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Solution

The correct option is C $l n = a m^{2}$
$l x + m y + n = 0$

So, $m y = - l x - n$

$y = - \frac{l}{m} x - \frac{n}{m}$

Comparing with $y = M x + c$ ,

$M = - \frac{l}{m}$

$c = - \frac{n}{m}$

We know that $c = \frac{a}{M}$

$M c = a$

Putting the values of M and c,

$\frac{- l}{m x} - \frac{n}{m} = a$

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

$l n = a m^{2}$

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Similar questions

l x + m y + n = 0

x^{2} = 4 a y

l x + m y + n = 0

y^{2} = 4 a x

l n = a m^{2}

y^{2} = 4 a x

l n = a m^{2} .

a x^{2} + 2 h x y + b y^{2} = 0

l x + m y + n = 0

\frac{n^{2} \sqrt{h^{2} - a b}}{∣ ∣ a m^{2} - 2 h l m + b l^{2} ∣ ∣}

a x^{2} + 2 h x y + b y^{2} = 0

l x + m y = 1

\frac{x^{'}}{l} = \frac{y^{'}}{m} = \frac{a + b}{a m^{2} - 2 h l m + b l^{2}} .

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