If the line $l x + m y + n = 0$ touches the parabola $y^{2} = 4 a x$ , then $l n =$

l n = a m

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l n = a m^{2}

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l n = \frac{1}{2} a m

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l n = a m^{3}

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Solution

The correct option is B $l n = a m^{2}$
The x-coordinates of the points of intersection of the line $l x + m y + n = 0$ or

$y = - (\frac{l x + n}{m})$ and the parabola $y^{2} = 4 a x$ are roots of the equation.

$[- (\frac{l x + n}{m})]^{2} = 4 a x$

$l^{2} x^{2} + 2 x (l n - 2 a m^{2}) + n^{2} = 0$

If the line $l x + m y + n = 0$ touches the parabola $y^{2} = 4 a x$ , then this equation has equal roots.

Therefore,

$4 (l n - 2 a m^{2})^{2} - 4 l^{2} n^{2} = 0$

$- 4 a l m^{2} n + 4 a^{2} m^{4} = 0$

$l n = a m^{2}$

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a_{1}, a_{2}, a_{3} . . . . a_{n}, . . . .

∣ ∣ ∣ ∣ \begin{matrix} ln a_{n} & ln a_{n + 1} & ln a_{n + 2} ln a_{n + 3} & ln a_{n + 4} & ln a_{n + 5} ln a_{n + 6} & ln a_{n + 7} & ln a_{n + 8} \end{matrix} ∣ ∣ ∣ ∣

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