Let one root of the quadratic equation in x: | vedclass

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(D) Let the roots of the quadratic equation be $\alpha$ and $2\alpha$.From the sum of roots, $\alpha + 2\alpha = 3\alpha = -\frac{9(k-1)}{k^2 - 15k +

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$12$

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(D) Let the roots of the quadratic equation be $\alpha$ and $2\alpha$.From the sum of roots, $\alpha + 2\alpha = 3\alpha = -\frac{9(k-1)}{k^2 - 15k + 27}$, which gives $\alpha = -\frac{3(k-1)}{k^2 - 15k + 27}$.From the product of roots, $\alpha(2\alpha) = 2\alpha^2 = \frac{18}{k^2 - 15k + 27}$.Substituting the value of $\alpha$ into the product equation: $2 \left[ -\frac{3(k-1)}{k^2 - 15k + 27} \right]^2 = \frac{18}{k^2 - 15k + 27}$.Simplifying: $\frac{18(k-1)^2}{(k^2 - 15k + 27)^2} = \frac{18}{k^2 - 15k + 27}$.This implies $(k-1)^2 = k^2 - 15k + 27$.Expanding the left side: $k^2 - 2k + 1 = k^2 - 15k + 27$.Solving for $k$: $13k = 26$, so $k = 2$.The parabola is $y^2 = 6kx$, which is $y^2 = 12x$.The length of the latus rectum of $y^2 = 4ax$ is $4a$. Here, $4a = 6k = 6(2) = 12$.

Let one root of the quadratic equation in $x$: $(k^2 - 15k + 27)x^2 + 9(k-1)x + 18 = 0$ be twice the other. Then the length of the latus rectum of the parabola $y^2 = 6kx$ is equal to:

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