Question

Find the equation of tangents to the curve `y=x^3` which are parallel to the line `12x-y-3=0`?

Answer 1

`12x-y-16=0` and `12x-y+16=0`

Explanation:

The line `12x-y-3=0` can be written as `y=12x-3` in point slope form and hence its slope is `12`. A line parallel to this line too will have slope `12`.

So we need to draw a tangent to `y=x^3`, which has slope as `12`.

As slope at a point is given by first derivative of the function i.e. `(dy)/(dx)` and

`(dy)/(dx)=3x^2`

we should have `3x^2=12` or `x^2-4=0` i.e. `x=2` or `-2`

Hence wewill have such a tangent at two points,

one at `(2,2^3)` i.e. `(2,8)` and at `(-2,(-2)^3)` i.e. `(-2,-8)`

and tangents will be

`y-8=12(x-2)` i.e. `12x-y-16=0` and

`y+8=12(x+2)` i.e. `12x-y+16=0`

graph{(y-x^3)(12x-y+16)(12x-y-16)(12x-y-3)=0 [-5, 5, -15, 15]}