Application of DerivativeHard
Question
The point (s) on the curve y3 + 3x2 = 12y where the tangent is vertical, is (are)
Options
A.
B.
C.(0, 0)
D.
Solution
Given, y3 + 3x2 = 12y ....(i)
⇒ 3y2
+ 6x = 12
⇒
⇒
For vertical tangent,
= 0
⇒ 12 - 3y2 = 0 ⇒ y =
2
On putting y = 2 in Eq. (i) we get x =
and again putting y = - 2 in Eq. (i) we get 3x2 = - 6, no real solution,
∴ The required point
⇒ 3y2
+ 6x = 12 ⇒
⇒
For vertical tangent,
= 0 ⇒ 12 - 3y2 = 0 ⇒ y =
2On putting y = 2 in Eq. (i) we get x =
and again putting y = - 2 in Eq. (i) we get 3x2 = - 6, no real solution, ∴ The required point
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