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

Answer:

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

Step-by-step explanation:

Given:

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

⇒ $x=9t^2+3$

then,

 $\frac{dy}{dt}=18t$

or,

⇒ $y=6t^3+3$

then,

 $\frac{dy}{dt}=18t^2$

⇒ $\frac{dy}{dx}=\frac{18t^2}{18t}$

        $=t$

By using the point slope form.

The equation of tangent will be:

⇒ $y-9=1(x-12)$

   $y-9=x-12$

          $y=x-12+9$

          $y=x-3$