parametric equation $x=2\ln (t+2)$and $y=t^3+2t+3$

Question

Given that the parametric equation $$x=2\ln (t+2)$$ $$y=t^3+2t+3$$

At the point P on the curve, the value of the parameter is p. It is given that the gradient of the curve at P is $\frac{1}{2}$. Show that $p=\frac{1}{3p^2+2}-2$

My attempt,

I found the $\frac{dy}{dx}$ which is $\frac{1}{2}(3t^2+2)(t+2)$

Then when I equal the $\frac{dy}{dx}=\frac{1}{2}$

I found this equation $3p^3+6p^2+2p+3=0$. I don't know how to simplify it to $p=\frac{1}{3p^2+2}-2$

Answer 1

Notice, $$3p^3+6p^2+2p+3=0$$ $$\underbrace{3p^3+2p}+6p^2+3=0$$ $$3p^3+2p=-6p^2-3$$ $$p(3p^2+2)=-6p^2-3$$ $$\implies p=\frac{-6p^2-3}{3p^2+2}$$$$=\frac{1-(6p^2+4)}{3p^2+2}$$ $$=\frac{1}{3p^2+2}-\frac{2(3p^2+2)}{3p^2+2}$$ $$=\color{blue}{\frac{1}{3p^2+2}-2}$$

Answer 2

$$3p^3+6p^2+2p+3=0 \\ \implies 3p^3+6p^2+2p+4=1 \\ \implies (p+2)(3p^2+2) = 1$$

Answer 3

You have $$\frac{dy}{dx}=\frac 12\Rightarrow 3p^2+2=\frac{1}{p+2}$$ $$\Rightarrow \frac{1}{3p^2+2}=p+2$$ and hence the result