Solving the Quadratic Equation for k

Given the quadratic equation: $(k^2 - 15k + 27)x^2 + 9(k - 1)x + 18 = 0$ Let the roots of this equation be $\\alpha$ and $2\alpha$, as one root is twice the other.

Using Vieta's formulas:

• Sum of roots: $\alpha + 2\alpha = 3\alpha = -\frac{9(k - 1)}{k^2 - 15k + 27}$
• Product of roots: $\alpha \cdot 2\alpha = 2\alpha^2 = \frac{18}{k^2 - 15k + 27}$

From the product of roots, we get $\alpha^2 = \frac{9}{k^2 - 15k + 27}$.

From the sum of roots, we get $\alpha = -\frac{3(k - 1)}{k^2 - 15k + 27}$. Squaring this gives $\alpha^2 = \frac{9(k - 1)^2}{(k^2 - 15k + 27)^2}$.

Equating the two expressions for $\alpha^2$ (assuming $k^2 - 15k + 27 \neq 0$): $ \frac{9}{k^2 - 15k + 27} = \frac{9(k - 1)^2}{(k^2 - 15k + 27)^2} $ $ 1 = \frac{(k - 1)^2}{k^2 - 15k + 27} $ $ k^2 - 15k + 27 = (k - 1)^2 $ $ k^2 - 15k + 27 = k^2 - 2k + 1 $ $ -15k + 27 = -2k + 1 $ $ 26 = 13k $ $ k = 2 $

We verify that for $k=2$, the coefficient of $x^2$ ($k^2 - 15k + 27 = 4 - 30 + 27 = 1$) and the coefficient of $x$ ($9(k-1) = 9(1) = 9$) are non-zero.

Calculating the Latus Rectum Length

The equation of the parabola is given as $y^2 = 6kx$. Substitute the value $k = 2$ into the parabola equation: $ y^2 = 6(2)x $ $ y^2 = 12x $

The standard form of a parabola opening to the right is $y^2 = 4ax$. Comparing $y^2 = 12x$ with the standard form $y^2 = 4ax$, we have: $ 4a = 12 $ $ a = 3 $

The length of the latus rectum of a parabola $y^2 = 4ax$ is $4a$. Therefore, the length of the latus rectum is $12$.