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Anusha Lalit Chana
Asked: 2022-11-09T05:15:01+05:30 In:

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)`

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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  1. 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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