If 2x + 3y + 12 = 0 and x - y + 4lambda = 0 a | vedclass

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(D) The condition for two lines $l_1x + m_1y + n_1 = 0$ and $l_2x + m_2y + n_2 = 0$ to be conjugate with respect to the parabola $y^2 = 4ax$ is given

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$-3$

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(D) The condition for two lines $l_1x + m_1y + n_1 = 0$ and $l_2x + m_2y + n_2 = 0$ to be conjugate with respect to the parabola $y^2 = 4ax$ is given by $l_1n_2 + l_2n_1 = 2am_1m_2$.Given the lines $2x + 3y + 12 = 0$ and $x - y + 4\lambda = 0$ and the parabola $y^2 = 8x$,we have $4a = 8$,so $a = 2$.Here,$l_1 = 2, m_1 = 3, n_1 = 12$ and $l_2 = 1, m_2 = -1, n_2 = 4\lambda$.Substituting these values into the condition:$2(4\lambda) + 1(12) = 2(2)(3)(-1)$$8\lambda + 12 = -12$$8\lambda = -24$$\lambda = -3$.

If $2x + 3y + 12 = 0$ and $x - y + 4\lambda = 0$ are conjugate with respect to the parabola $y^2 = 8x$,then $\lambda$ is equal to

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