Q. 134.5( 2 Votes )
Find a point on the curve y = 3x2 + 4 at which the tangent is perpendicular to the line whose slope is 
Answer :
Given:
The curve y = 3x2 + 4 and the Slope of the tangent is
y = 3x2 + 4
Differentiating the above w.r.t x
⇒ = 2
3x2 – 1 + 0
⇒ = 6x ...(1)
Since, tangent is perpendicular to the line,
The Slope of the normal =
i.e, =
⇒ =
⇒ x = 1
Substituting x = 1 in y = 3x2 + 4,
⇒ y = 3(1)2 + 4
⇒ y = 3 + 4
⇒ y = 7
Thus, the required point is (1,7).
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