JEE Main 2026 · 6 Apr · Shift 1 · Mathematics Q2

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MathematicsSingle correct+4 / -1Question 2 of 75

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

Answer and worked solution

Official answer: D

Roots r and 2r. Let $A=k^2-15k+27$ .

Sum: $3r=-\dfrac{9(k-1)}A$ , so $r=-\dfrac{3(k-1)}A$ .
Product: $2r^2=\dfrac{18}A$ , which gives $\dfrac{18(k-1)^2}{A^2}=\dfrac{18}A$ .

So $(k-1)^2=A$ : $k^2-2k+1=k^2-15k+27$ , and $k=2$ .

Latus rectum of $y^2=12x$ : 12.

(D)

Topics:Roots and coefficients Parabola

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JEE Main 2026 · 6 Apr · Shift 1, Mathematics question 2