JEE Mains · Maths · STD 12 - 6. Application of derivatives
Let the area enclosed by the \(x\)-axis, and the tangent and normal drawn to the curve \(4 x ^{3}-\) \(3 x y^{2}+6 x^{2}-5 x y-8 y^{2}+9 x+14=0\) at the point \((-2,3)\) be \(A\). Then \(8 A\) is equal to \(.......\)
- A \(174\)
- B \(132\)
- C \(185\)
- D \(170\)
Answer & Solution
Correct Answer
(D) \(170\)
Step-by-step Solution
Detailed explanation
\(4 x^{3}-3 x y^{2}+6 x^{2}-5 x y-8 y^{2}+9 x+14=0\) differentiating both sides we get \(12 x^{2}-3 y^{2}-6 x y y^{\prime}+12 x-5 y-5 x y^{\prime}-16 y y^{\prime}+9=0\) \(18-27+36 y^{\prime}-21-15+10 y^{\prime}-18 y^{\prime}+9=0\) \(2 y^{\prime}=-9\)…
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