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?

Eliminating the parameter $t$ from the first equation yields
$$x=(y-1)^2 $$
giving 
$$\frac{dx}{dy}=2(y-1)\text{ or }\frac{dy}{dx}=\frac{1}{2(y-1)}$$
which gives $\dfrac{dx}{dy}$ a value $0$ at $(0,1)$. So the first curve has a vertical tangent at $(0,1)$ and a slope of $\dfrac{dy}{dx}=-\frac{1}{2}$ at $(0,1)$.
For the second equation we may let
$$ F(x,y)=5x^2+5xy+3y^2-8x-6y+3 $$
and find the implicit derivative using the partial derivative identity
$$ \frac{dy}{dx}=-\frac{F_x}{F_y} $$
to obtain
$$ \frac{dy}{dx}=-\frac{10x+5y-8}{5x+6y-6} $$
alternately
$$ \frac{dx}{dy}=-\frac{5x+6y-6}{10x+5y-8} $$
This shows that at the point $(1,0)$ we have
$$ \frac{dy}{dx}=2 $$
and at the point $(0,1)$ we have a vertical tangent since $\dfrac{dx}{dy}=0$ at that point.
Thus at the point $(0,1)$ both curves have a vertical tangent. So the angle between their tangents is $0$ at $(0,1)$.
At the point $(1,0)$ the slopes are the negative reciprocal of each other so the angle between their tangents is $90^\circ$ or $\pi/2$ radians at that point.

