If the lines 2x + 3y + 12 = 0 and x - y + k = | vedclass

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(C) Given,the conjugate lines are $2x + 3y + 12 = 0$ $(i)$ and $x - y + k = 0$ $(ii)$.Two lines are conjugate with respect to a parabola if the pole o

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$-12$

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(C) Given,the conjugate lines are $2x + 3y + 12 = 0$ $(i)$ and $x - y + k = 0$ $(ii)$.Two lines are conjugate with respect to a parabola if the pole of one line lies on the other line.Let the pole of the line $2x + 3y + 12 = 0$ be $(x_1, y_1)$ with respect to the parabola $y^2 = 8x$.The equation of the polar of the point $(x_1, y_1)$ with respect to $y^2 = 8x$ is $yy_1 = 4(x + x_1)$,which simplifies to $4x - yy_1 + 4x_1 = 0$.Comparing this with the given line $2x + 3y + 12 = 0$,we have:$\frac{4}{2} = \frac{-y_1}{3} = \frac{4x_1}{12}$$2 = \frac{-y_1}{3} \Rightarrow y_1 = -6$$2 = \frac{x_1}{3} \Rightarrow x_1 = 6$Thus,the pole is $(6, -6)$.Since the lines are conjugate,the pole $(6, -6)$ must lie on the second line $x - y + k = 0$.Substituting the coordinates,we get $6 - (-6) + k = 0$.$12 + k = 0 \Rightarrow k = -12$.

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

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