# Find a point on t

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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