Hello Tweety Originally Posted by

**Tweety** The curve with parametric equations $\displaystyle x = 3t^{2} , y = 4t^{3} $ along with the tangent at the point with parameter t = 1

find the equation if this tangent and the coordinates of the point where it meets the curve again.

$\displaystyle \frac{dx}{dt} = 6t $

$\displaystyle \frac{dy}{dt} = 12t^{2} $

$\displaystyle \frac{dy}{dx} = \frac{12t^{2}}{6t} = 2t $

when t = 1

$\displaystyle x = 3 , y = 4 , \frac{dy}{dx} = 2 $

equation of tangent = $\displaystyle y -4 = 2(x-3) $

$\displaystyle y = 2x-2 $

however to find the coordinates of where the curve crosses the tangent again, would I not just be able to sub in the values of x and y into the tnagent equation?

$\displaystyle 4t^{3} = 2(3t^{2})-2 $

$\displaystyle 4t^{3} = 6t^{2} - 2 $ can't seem to solve from here,

any help appreicated.

thanks

Your working is perfect up to here. Good job!

What you need to realise now is that the tangent at $\displaystyle t=1$ effectively cuts the curve in *two coincident points *where $\displaystyle t = 1$.

So $\displaystyle t=1$ will be a *repeated root *of your final equation, which is more simply written as:

$\displaystyle 2t^3-3t^2+1=0$

In other words $\displaystyle (t-1)^2$, or $\displaystyle (t^2-2t+1)$, is a factor of the LHS of:$\displaystyle 2t^3 - 3t^2 +1 =0$

So we get:$\displaystyle (t^2-2t+1)(2t+1)=0$

(Check it out: multiply out the brackets and you get $\displaystyle 2t^3-3t^2+1$.)

So the other value of $\displaystyle t$ is ...?

I'm sure you can take it from here.

Grandad