Finding intersection angle at intersection point of two curves

Question

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?

Answer 1

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.

Answer 2

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?

Answer 3

Hint:

The "angle" between the two curves at the point of intersection is defined as the angle between the tangent vectors of the two curves at that point of intersection.

Note that if $\frac{dy}{dx}$ is the slope of the tangent, then $\langle 1, \frac{dy}{dx} \rangle$ is a tangent vector.

Answer 4

The "angle between the curves" at a given point is the angle between the tangent lines to the curves at that point.

The tangent vector to the line given by $x= t^2$, $y= t+ 1$ at any t is <2t, 1>. For the curve given by $5x^2+ 5xy+ 3y^2- 8x- 6y+ 3= 0$, $10x+ 5y+ 4xy'+ 5yy'- 8- 5y'= (4x+ 5y- 5)y'+ 10x+ 5y- 8= 0$ so $y'= \frac{10x+ 5y- 8}{4x+ 5y- 5}$.

At (1, 0), the tangent vector for the first curve is <2, 1>. For the second curve $y'= \frac{10}{4}= \frac{5}{2}$ so a tangent vector is <2, 5>. Use the fact that $u\cdot v= |u||v| cos(\theta)$ where $\theta$ is the angle between the vectors.