Latus Rectum of Parabola from Quadratic Equation Roots

To find the length of the latus rectum of the parabola $y^2 = 6kx$, we first need to determine the value of $k$ using the given quadratic equation:

$(k^2 - 15k + 27)x^2 + 9(k - 1)x + 18 = 0$

Let the roots of the quadratic equation be $\alpha$ and $2\alpha$.

Using the relation between the roots and coefficients of a quadratic equation:

1. Sum of roots: $\alpha + 2\alpha = -\frac{9(k - 1)}{k^2 - 15k + 27}$ $3\alpha = -\frac{9(k - 1)}{k^2 - 15k + 27}$ $\alpha = -\frac{3(k - 1)}{k^2 - 15k + 27} \quad \text{--- (1)}$

2. Product of roots: $\alpha \cdot 2\alpha = \frac{18}{k^2 - 15k + 27}$ $2\alpha^2 = \frac{18}{k^2 - 15k + 27}$ $\alpha^2 = \frac{9}{k^2 - 15k + 27} \quad \text{--- (2)}$

Now, squaring equation (1) and equating it to equation (2): $\left(-\frac{3(k - 1)}{k^2 - 15k + 27}\right)^2 = \frac{9}{k^2 - 15k + 27}$

$\frac{9(k - 1)^2}{(k^2 - 15k + 27)^2} = \frac{9}{k^2 - 15k + 27}$

Since $k^2 - 15k + 27 \neq 0$, we can cancel $9(k^2 - 15k + 27)$ from both sides: $(k - 1)^2 = k^2 - 15k + 27$

Expanding the left side: $k^2 - 2k + 1 = k^2 - 15k + 27$

$-2k + 1 = -15k + 27$

$13k = 26$ $k = 2$

Now, for the parabola $y^2 = 6kx$, the standard equation is $y^2 = 4ax$, where the length of the latus rectum is the coefficient of $x$, which is $6k$.

Substituting $k = 2$: $\text{Length of latus rectum} = 6k = 6 \times 2 = 12$

Thus, the length of the latus rectum of the parabola is equal to 12.