JEE Mains · Maths · STD 12 - 8. Application and integration
Let \(T\) and \(C\) respectively be the transverse and conjugate axes of the hyperbola \(16 x^2-\) \(y^2+64 x+4 y+44=0\). Then the area of the region above the parabola \(x^2=y+4\), below the transverse axis \(T\) and on the right of the conjugate axis \(C\) is:
- A \(4 \sqrt{6}+\frac{44}{3}\)
- B \(4 \sqrt{6}+\frac{28}{3}\)
- C \(4 \sqrt{6}-\frac{44}{3}\)
- D \(4 \sqrt{6}-\frac{28}{3}\)
Answer & Solution
Correct Answer
(B) \(4 \sqrt{6}+\frac{28}{3}\)
Step-by-step Solution
Detailed explanation
\(16\left(x^2+4 x\right)-\left(y^2-4 y\right)+44=0\) \(16(x+2)^2-64-(y-2)^2+4+44=0\) \(16(x+2)^2-(y-2)^2=16\) \(\frac{(x+2)^2}{1}-\frac{(y-2)^2}{16}=1\) \(A=\int \limits_{-2}^{\sqrt{6}}\left(2-\left(x^2-4\right)\right) d x\)…
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