# 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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