If the lines 2x - y + 3 = 0 and 4x + ky + 3 = | vedclass

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(D) The given lines are $2x - y + 3 = 0$ and $4x + ky + 3 = 0$. For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,two lines $a_1x + b_1y + c_1 =

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$6$

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(D) The given lines are $2x - y + 3 = 0$ and $4x + ky + 3 = 0$. For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,two lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ are conjugate if $a^2a_1a_2 + b^2b_1b_2 = c_1c_2$. Rewriting the ellipse equation $5x^2 + 6y^2 = 15$ as $\frac{x^2}{3} + \frac{y^2}{2.5} = 1$,we have $a^2 = 3$ and $b^2 = 2.5 = \frac{5}{2}$. Here,$a_1 = 2, b_1 = -1, c_1 = 3$ and $a_2 = 4, b_2 = k, c_2 = 3$. Substituting these values into the condition: $3(2)(4) + (2.5)(-1)(k) = (3)(3)$. $24 - 2.5k = 9$. $2.5k = 15$. $k = \frac{15}{2.5} = 6$.

If the lines $2x - y + 3 = 0$ and $4x + ky + 3 = 0$ are conjugate with respect to the ellipse $5x^2 + 6y^2 - 15 = 0$,then $k$ equals:

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