AP EAMCET · Maths · Area Under Curves
The area of the region (in sq.units) bounded by the curves \(x^2+y^2=16\) and \(\mathrm{y}^2=6 \mathrm{x}\) is
- A \(4 \pi+4 \sqrt{3}\)
- B \(\frac{2}{3}(4 \pi+\sqrt{3})\)
- C \(\frac{4}{3}(4 \pi+\sqrt{3})\)
- D \(\frac{4 \pi+\sqrt{3}}{3}\)
Answer & Solution
Correct Answer
(C) \(\frac{4}{3}(4 \pi+\sqrt{3})\)
Step-by-step Solution
Detailed explanation
Solve \(x^2+y^2=16\) and \(y^2=6x\): \(x^2+6x=16 \implies x^2+6x-16=0 \implies (x-2)(x+8)=0 \implies x=2\) (since \(x \ge 0\) for \(y^2=6x\)). \(y^2=6(2)=12 \implies y=\pm 2\sqrt{3}\). Intersection points: \((2, \pm 2\sqrt{3})\). Area…
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