JEE Main 2022 Mathematics question
Let the area enclosed by the x-axis, and the tangent and normal drawn to the curve at the point (−2, 3) be A. Then 8A is equal to
Answer: 170
Solution
Differentiating the equation of the curve implicitly and putting (−2, 3) gives = −9, so the slope of the tangent is −9/2 and of the normal is 2/9. The tangent y − 3 = −(9/2)(x + 2) meets y = 0
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