JEE Mains · Maths · STD 12 - 8. Application and integration
If the area of the region bounded by \(16x^2 - 9y^2 = 144\) and \(8x - 3y = 24\) is A, then \(3(A + 6 \log_e(3))\) is equal to _______.
- A 20
- B 22
- C 24
- D 26
Answer & Solution
Correct Answer
(C) 24
Step-by-step Solution
Detailed explanation
The given equations are: Hyperbola: \(16x^2 - 9y^2 = 144 \Rightarrow \dfrac{x^2}{9} - \dfrac{y^2}{16} = 1\) Line: \(8x - 3y = 24 \Rightarrow y = \dfrac{8}{3}(x - 3)\) To find the points of intersection, substitute \(y\) from the line equation into the hyperbola equation:…
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