Find the point on the curve `y=3x^2+4` at which the tangent is perpendicular to the line whose slope is `-1/6` .

Answer

Point is `(1,7)`.

Explanation

Let US consider the points be `(x_{1}, y_{1})`.

Slope of the line `=\frac{-1}{6}`


`\therefore` Slope of the line perpendicular to it `=6`

As point lies on the curve.

Hence, `y_{1}=3 x_{1}^{2}+4`

Now, `y=3 x^{2}+4`
`\therefore \frac{d y}{d x}=6 x`


Now,


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