JEE Main 2026 — 6 April, Morning Shift — Question 22

Let the roots be $\alpha$ and $2\alpha$. For the quadratic $(k^2 - 15k + 27)x^2 + 9(k-1)x + 18 = 0$, 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, $\alpha^2 = \frac{9}{k^2 - 15k + 27}$. From the sum, $\alpha = -\frac{3(k-1)}{k^2 - 15k + 27}$. Square the sum expression: $\alpha^2 = \frac{9(k-1)^2}{(k^2 - 15k + 27)^2}$. Equate the two expressions for $\alpha^2$: $\frac{9}{k^2 - 15k + 27} = \frac{9(k-1)^2}{(k^2 - 15k + 27)^2}$. Cancel 9 and multiply: $k^2 - 15k + 27 = (k-1)^2 = k^2 - 2k + 1$. Simplify: $-15k + 27 = -2k + 1 \Rightarrow -13k = -26 \Rightarrow k = 2$. The parabola is $y^2 = 6kx = 12x$, so length of latus rectum = $12$.