JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
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:
- A \(4\)
- B \(6\)
- C \(8\)
- D \(12\)
Answer & Solution
Correct Answer
(D) \(12\)
Step-by-step Solution
Detailed explanation
Let the roots of the given quadratic equation be \(\alpha\) and \(2\alpha\). Sum of the roots: \(\alpha + 2\alpha = \dfrac{-9(k-1)}{k^2 - 15k + 27}\) \(3\alpha = \dfrac{-9(k-1)}{k^2 - 15k + 27} \Rightarrow \alpha = \dfrac{-3(k-1)}{k^2 - 15k + 27}\) Product of the roots:…
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