Maths Q2 – JEE-MAIN 2026

Home Maths Complex & QuadraticQuestion #82

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:

Clue 1 of 3: Roots of Quadratic Equation

When roots of a quadratic equation have a specific relationship (e.g., one is twice the other), Vieta's formulas are crucial for establishing relationships between the coefficients and the roots.

🧠 Full Solution Path

Step 1: Apply Vieta's formulas for the given root relationship✦ Active

Let the roots of the quadratic equation $(k^{2} - 15 k + 27) x^{2} + 9 (k - 1) x + 18 = 0$ be $α$ and $2 α$ . Using Vieta's formulas:

3 α = - \frac{9 (k - 1)}{k^{2} - 15 k + 27} \Rightarrow α = - \frac{3 (k - 1)}{k^{2} - 15 k + 27} (1)

2 α^{2} = \frac{18}{k^{2} - 15 k + 27} \Rightarrow α^{2} = \frac{9}{k^{2} - 15 k + 27} (2)

Step 2: Solve for the value of k○ Expand

Square equation (1) and equate it to equation (2) to eliminate $α$ :

{(- \frac{3 (k - 1)}{k^{2} - 15 k + 27})}^{2} = \frac{9}{k^{2} - 15 k + 27}

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

Assuming $k^{2} - 15 k + 27 \neq 0$ , we can simplify:

(k - 1)^{2} = k^{2} - 15 k + 27

k^{2} - 2 k + 1 = k^{2} - 15 k + 27

13 k = 26 \Rightarrow k = 2

For $k = 2$ , $k^{2} - 15 k + 27 = 4 - 30 + 27 = 1 \neq 0$ , so $k = 2$ is a valid solution.

Step 3: Calculate the length of the latus rectum○ Expand

Substitute $k = 2$ into the parabola equation $y^{2} = 6 k x$ :

y^{2} = 6 (2) x \Rightarrow y^{2} = 12 x

Comparing this with the standard form of a parabola $y^{2} = 4 a x$ , the length of the latus rectum is $4 a$ . From $y^{2} = 12 x$ , we have $4 a = 12$ .

Length of latus rectum = 12

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Maths Question 2 – JEE-MAIN 2026

Clue 1 of 3: Roots of Quadratic Equation

🧠 Full Solution Path