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The point(s) on thecurve `y^3+ 3x^2=12 y`where the tangent isvertical, is(are) ??`(+-4/(sqrt(3)), -2)`(b) `(+- sqrt((11)/3, ) 1)``(0, 0)`(d) `(+-4/(sqrt(3)), 2)`A. `(pm(4)/(sqrt(3)),-2)`B. `(pmsqrt((pi)/(3)),1)`C. (0, 0)D. `(pm(4)/(sqrt(3)),2)` |
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Answer» Correct Answer - D We have, `y^(3)+3x^(2)=12y " …(i)" ` `rArr 3y^(2) (dy)/(dx)+6x=12(dy)/(dx) " " `[Diff. w.r.t. x] `rArr 3(y^(2)-4) (dy)/(dx)= -6x rArr (dy)/(dx) = -(2x)/(y^(2)-4)` At point(s) where the tangent(s) is (are) vertical, `(dy)/(dx)` is not defined . `therefore y^(2) -4=0 rArr y= pm 2`. From (i), we find that `y=2 rArr x=pm(4)/(sqrt(3))` and, `y = -2 rArr x^(2) =(-16)/(3)`, which is not possible. Hence, the required points are `(pm 4//sqrt(3), 2)` |
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