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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)/(sqrt3), -2)`B. `(pm sqrt((11)/(3)), 0)`C. `(0, 0`)D. ` (pm (4)/(sqrt3), 2)` |
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Answer» Correct Answer - D Given, `" " y^(3) + 3x ^(2) = 12 y " " `…(i) On differentiating w.r.t. x, we get `rArr " " 3y^(2) (dy)/(dx) + 6x = 12 (dy)/(dx)` `rArr " " (dy)/(dx) = ( 6x)/(12 - 3y^(2))` `rArr " " (dx)/(dy) = ( 12 - 3y^(2))/(6x)` For vertical tangent, `(dx)/(dy) =0` `rArr 12 - 3y^(2) =0 rArr y = pm 2` On putting, `y = 2` in Eq. (i), we get ` x = pm ( 4)/(sqrt3)` and again putting `y = - 2 ` in Eq. (i), we get ` 3x^(2) = - 16`, no real solution . So, the required point is `(pm (4)/(sqrt(3)), 2)`. |
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