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If $2 x - y + 3 = 0, 4 x + k y + 3 = 0$ are conjugate with respect to the ellipse $5 x^{2} + 6 y^{2} - 15 = 0$ then $k =$

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answer is D.

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Detailed Solution

$\frac{x^{2}}{3} + \frac{2 y^{2}}{5} = 1$

$a^{2} l_{1} l_{2} + b^{2} m_{1} m_{2} = n_{1} n_{2}$

$3 \times 8 + \frac{5}{2} \times (- k) = 9$

$24 - \frac{5 k}{2} = 9$

$\Rightarrow 24 - 9 = \frac{5 k}{2}$

$∴ k = 6$

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