Determining the angle between two intercepting curves I'm having trouble solving the following problem:
So i have the following two curves 
$(x,y) = (t^2, t+1)$,  and $ 5x^2 +5xy +3y^2 - 8x-6y+3=0$
In the first part of the problem i'm asked to find the intercept point. I did this by inserting $(x,y) = (t^2, t+1)$ in the second curve and solving for t. 
Long story short, i get that the intercept points are (0.1) and (1.0). 
However in the second part of the problem im asked to find the angle between the curves at the intercepting points.
Im aware that i find the angle between the curves by first determining their respective gradient at the point. However this is where i run into trouble. I know how to determine the gradient for the second curve. But for the first one, since its written on parametic form, i have no clue how to determine the gradient. 
 A: This is how you determine the gradient:
Please note that $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
In this case, $\frac{dy}{dx} = \frac{1}{2t}$
Hope you can solve the problem now!
A: Since $$x=(y-1)^2,$$ we have $$1=2(y-1)y',$$ which in the point $(1,0)$ gives a slop $m_1=-\frac{1}{2}$
Now, from $$ (5x^2 +5xy +3y^2 - 8x-6y+3)'=0$$ take a second slop.
Can you end it now?
In this case with the point $(1,0)$ I got $90^{\circ}.$
A: You only need a vector parallel to the gradient. It’s not important to get its length right since you’re going to normalize it, anyway, to compute the angle.  
Now, the gradient is perpendicular to the tangent to a curve. So, if $(x,y)=(t^2,t+1)$, the tangent vector at $t$ is $(2t,1)$, and a vector perpendicular to this is $(-1,2t)$.
A: The given curves are: $$x=t^2, y=t+1$$$$\implies x=(y-1)^2$$
$$\implies \frac{dy}{dx}=\frac{1}{2(y-1)}$$ &
$$5x^2+5xy+3y^2-8x-6y+3=0$$
$$\implies \frac{dy}{dx}=\frac{8-10x-5y}{5x+6y-6}$$
Now, solving above two equations of curves, we get two distinct points of intersection as $(0, 1)$ & $(1, 0)$
Now, computing slope of tangent to the curve: $x=(y-1)^2$ at the point $(0, 1)$ as follows
$$m_1=\left(\frac{dy}{dx}\right)_{(0, 1)}=\left(\frac{1}{2(y-1)}\right)_{(0, 1)}=\infty$$
Similarly, computing slope of tangent to the curve: $5x^2+5xy+3y^2-8x-6y+3=0$ at the point $(0, 1)$ as follows
$$m_2=\left(\frac{dy}{dx}\right)_{(0, 1)}=\left(\frac{8-10x-5y}{5x+6y-6}\right)_{(0, 1)}=\infty$$
Since both the tangents are vertical (since their slope is $\infty$) so the angles between them is zero i.e. the curves are touching one another at the point $(0, 1)$
Computing slope of tangent to the curve: $x=(y-1)^2$ at the point $(1, 0)$ as follows
$$m_1=\left(\frac{dy}{dx}\right)_{(1, 0)}=\left(\frac{1}{2(y-1)}\right)_{(1, 0)}=-1$$
Similarly, computing slope of tangent to the curve: $5x^2+5xy+3y^2-8x-6y+3=0$ at the point $(1, 0)$ as follows
$$m_2=\left(\frac{dy}{dx}\right)_{(1, 0)}=\left(\frac{8-10x-5y}{5x+6y-6}\right)_{(1, 0)}=2$$
Now, the angle ($\theta$) between tangents to the curves is given as
$$\tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right|$$
$$\tan\theta=\left|\frac{-1-2}{1+(-1)(2)}\right|$$
$$\tan\theta=3$$
$$\theta=\tan^{-1}3=71.565^\circ$$
