Find a point on the curve y = 3x2 + 4 at which the tangent is perpendicular to the line whose slope is

Given:


The curve y = 3x2 + 4 and the Slope of the tangent is


y = 3x2 + 4


Differentiating the above w.r.t x


⇒ = 23x2 – 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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