Question

How do you find the equation of the tangent line to the curve `y=1+x^3` that is parallel to the line `12x-y=1`?

Answer 1

`12x-y-1=0`

Explanation:

If the tangent is parallel to `12x-y=1`, they must both have the same gradient.

Rearranging the eq of the line to gradient-intercept form,

`y=12x-1`

`:.` gradient`=12`

The first derivative of the curve would be the gradient of the curve, as well as the gradient of tangents to the curve at different points all along the curve.

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

At the point on the curve where the gradient of the tangent`=12`, `(dy)/(dx)=12`

`:.3x^2=12`

`x^2=4`

`x=+-2`

When `x=2`, `y=12(2)-1=23`

When `x=-2`, `y=12(-2)-1=-25`

Using the point-gradient formula, `y-y_1=m(x-x_1),`

`y-23=12(x-2)` or `y+25=12(x+2)`

`12x-y-1=0` or `12x-y-1=0`, which are the same lines.

`:.` tangent: `12x-y-1=0`, which is incidentally the line which we started with.