# parametric equations help

#### 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

MHF Hall of Honor
Hello 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.

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