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

Explanation

Given curve is `y=3x^2+4`
We have to seek out the point on the curve `y=3x^2+4`
at which the tangent is perpendicular to the line whose slope is `-1/6`
Let `(x,y)` be the points.
Slop of the given line =`-1/6`
∴ Slop of the point perpendicular to is `= 6`
`y=3x^2+4`
Differentiate w.r.t `x` we get ,
`=>(dy)/(dx)=6x`
Slop of the tan gent at `(x,y)=(dy)/(dx)=6x`
Slope of the tangent at `(x,y)=` Slop of the given line.
`=>6x=6`
`=>x=1`
Now,
`=>y=3x^2+4`
`=3(1)^2+4`
`=7`
∴ The required point is `(1,7)`.