JEE Mains · Maths · STD 12 - 8. Application and integration
The area of the region bounded by \(y^{2}=8 x\) and \(y^{2}=16(3-x)\) is equal to
- A \(\frac{32}{3}\)
- B \(\frac{40}{3}\)
- C \(16\)
- D \(19\)
Answer & Solution
Correct Answer
(C) \(16\)
Step-by-step Solution
Detailed explanation
finding their intersection \(pts\) \(y^{2}=8 x\; and\; y^{2}=-16(x-3)\) \(8 x=-16 x+48\) \(24 x=48\) \(x=2 ; y=\pm 4\) Required Area \(=2 \cdot \int \limits_{0}^{4}({3-\frac{y^{2}}{16}}-{\frac{y^{2}}{8}}) d y\)…
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