Mathematics

Mathematics

Problem

The point(s) on the curve y3+3x2=12yy^3 + 3x^2 = 12y , where the tangent is vertical, is (are) (a) (±43,−2)\left(\pm \frac{4}{\sqrt{3}}, -2\right) (b) (±113,1)\left(\pm \sqrt{\frac{11}{3}}, 1\right) (c) (0,0)(0, 0) (d) (±43,2)\left(\pm \frac{4}{\sqrt{3}}, 2\right)

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