JEE Main 2026 Apr 6 Shift 1, Mathematics Q2: Quadratic Equations

JEE Main2026Apr 6, Shift 1Mathematics

Let one root of the quadratic equation in $x$ : $(k^{2} - 15 k + 27) x^{2} + 9 (k - 1) x + 18 = 0$ be twice the other. Then the length of the latus rectum of the parabola $y^{2} = 6 k x$ is equal to:

A
$4$
B
$6$
C
$8$
D
$12$

Solution

Let the roots be $α$ and $2 α$ . By Vieta's relations:

Sum: $3 α = \frac{- 9 ( k - 1 )}{k ^{2} - 15 k + 27}$ .

Product: $2 α^{2} = \frac{18}{k ^{2} - 15 k + 27}$ .

From the sum, $α = \frac{- 3 ( k - 1 )}{k ^{2} - 15 k + 27}$ . Squaring and substituting into the product relation gives $\frac{9 ( k - 1 ) ^{2}}{( k ^{2} - 15 k + 27 ) ^{2}} = \frac{9}{k ^{2} - 15 k + 27}$ , i.e. $(k - 1)^{2} = k^{2} - 15 k + 27$ .

Expanding, $k^{2} - 2 k + 1 = k^{2} - 15 k + 27 \Rightarrow 13 k = 26 \Rightarrow k = 2$ .

Hence the parabola is $y^{2} = 12 x$ , whose latus rectum has length $12$ .

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JEE Main 2026 Apr 6 Shift 1, Mathematics Q2: Quadratic Equations

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