Finding intersection angle at intersection point of two curves I've got two curves:
$$(x,y) = (t^2,t+1), \quad t\in\mathbb{R}$$
$$5x^2 + 5xy + 3y^2 -8x -6y + 3 = 0$$
I've found the intersection points: 
$$(0,1) , (1,0)$$. 
But I can't figure out how to the the angle between the two curves at these intersection points? Should I use derivative somehow? How do I derive a parametric function?

plugging $$x=t^2,y=t+1$$ in the given equation we get
$$5t^4+5t^2(t+1)+3(t+1)^2-8t^2-6(t+1)+3=0$$
which can be simplified to $$5t^3(t+1)=0$$
can you solve it?
