Question

Find the equations of the tangents to the curve x=9t2+3, y=6t3+3 that pass through the point (12,9).

Answer 1

The equation will be "
`y=x-3`".

Explanation:

Given:

Points (12, 9) = (x, y)

⇒
`x=9t^2+3`

then,

`(dy)/(dt)=18t`

or,

⇒
`y=6t^3+3`

then,

`(dy)/(dt)=18t^2`

⇒
`(dy)/(dx)=(18t^2)/(18t)`

`=t`

By using the point slope form.

The equation of tangent will be:

⇒
`y-9=1(x-12)`

`y-9=x-12`

`y=x-12+9`

`y=x-3`