JEE Main 2026 — Mathematics Algebra

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

Sum of the roots:

$$\alpha + 2\alpha = \dfrac{-9(k-1)}{k^2 - 15k + 27}$$α+2α=k2−15k+27−9(k−1)​

$$3\alpha = \dfrac{-9(k-1)}{k^2 - 15k + 27} \Rightarrow \alpha = \dfrac{-3(k-1)}{k^2 - 15k + 27}$$3α=k2−15k+27−9(k−1)​⇒α=k2−15k+27−3(k−1)​

Product of the roots:

$$\alpha \cdot 2\alpha = \dfrac{18}{k^2 - 15k + 27}$$α⋅2α=k2−15k+2718​

$$2\alpha^2 = \dfrac{18}{k^2 - 15k + 27} \Rightarrow \alpha^2 = \dfrac{9}{k^2 - 15k + 27}$$2α2=k2−15k+2718​⇒α2=k2−15k+279​

Substituting the value of $\alpha$α from the sum into the product equation:

$$\left( \dfrac{-3(k-1)}{k^2 - 15k + 27} \right)^2 = \dfrac{9}{k^2 - 15k + 27}$$(k2−15k+27−3(k−1)​)2=k2−15k+279​

$$\dfrac{9(k-1)^2}{(k^2 - 15k + 27)^2} = \dfrac{9}{k^2 - 15k + 27}$$(k2−15k+27)29(k−1)2​=k2−15k+279​

Since $k^2 - 15k + 27 \neq 0$k2−15k+27=0, we can simplify to:

$$(k-1)^2 = k^2 - 15k + 27$$(k−1)2=k2−15k+27

$$k^2 - 2k + 1 = k^2 - 15k + 27$$k2−2k+1=k2−15k+27

$$13k = 26 \Rightarrow k = 2$$13k=26⇒k=2

The equation of the parabola is $y^2 = 6kx$y2=6kx. Substituting $k = 2$k=2, we get:

$$y^2 = 12x$$y2=12x

The length of the latus rectum of a parabola $y^2 = 4ax$y2=4ax is $4a$4a, which is the coefficient of $x$x.

Therefore, the length of the latus rectum is $12$12.

Answer: $12$12