JEE Mains · Maths · STD 12 - 8. Application and integration
If the area (in sq. units) bounded by the parabola \(y^2 =4\lambda x\) and the line \(y = \lambda x\), \(\lambda > 0\), is \(\frac{1}{9}\), then \(\lambda \) is equal to
- A \(48\)
- B \(4\sqrt 3\)
- C \(2\sqrt 6\)
- D \(24\)
Answer & Solution
Correct Answer
(D) \(24\)
Step-by-step Solution
Detailed explanation
\(y^{2}=4 \lambda x\) and \(y=\lambda x\) \(\lambda^{2} x^{2}=4 \lambda x\) \(x=0\) and \(x=\frac{4}{\lambda}\) Area \( = \int\limits_0^{4\lambda } {(\sqrt {4\lambda x} - \lambda x)} dx = \frac{1}{9}\)…
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