JEE Main 2026 — Quadratic Equation Question with Solution

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

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

Product of the roots:
$\alpha \cdot 2\alpha =\frac{18}{k^2-15k+27}$
$2{\alpha }^2=\frac{18}{k^2-15k+27}\Rightarrow {\alpha }^2=\frac{9}{k^2-15k+27}$

Substituting the value of $\alpha$ from the sum into the product equation:
${\left(\frac{-3(k-1)}{k^2-15k+27}\right)}^2=\frac{9}{k^2-15k+27}$

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

Since $k^2-15k+27\ne 0$, we can simplify to:
$(k-1{)}^2=k^2-15k+27$

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

$13k=26\Rightarrow k=2$

The equation of the parabola is $y^2=6kx$. Substituting $k=2$, we get:
$y^2=12x$

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

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

Answer: $12$