A) \[\left( \pm \frac{4}{\sqrt{3}},-2 \right)\]
B) \[\left( \pm \sqrt{\frac{11}{3}},0 \right)\]
C) \[(0,0)\]
D) \[\left( \pm \frac{4}{\sqrt{3}},2 \right)\]
Correct Answer: D
Solution :
Given, y3 + 3x2 = 12y ...(i) On differentiating w.r.t. x, we get \[3{{y}^{2}}\frac{dy}{dx}+6x=12\frac{dy}{dx}\]\[\Rightarrow \]\[\frac{dy}{dx}=\frac{6x}{12-3{{y}^{2}}}\] \[\Rightarrow \]\[\frac{dx}{dy}=\frac{12-3{{y}^{2}}}{6x}\] For vertical tangent, \[\frac{dx}{dy}=0\] \[\Rightarrow \]\[12-3{{y}^{2}}=0\]\[\Rightarrow \]\[y=\pm 2\] On putting y = 2 in Eq. (i), we get \[x=\pm \frac{4}{\sqrt{3}}\]and on putting y = - 2 in Eq. (i), we get 3x2 = -16, no real solution. \[\therefore \]The required points are \[\left( \pm \frac{4}{\sqrt{3}},2 \right).\]
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