Parameterize $3x^2 + 4xy + 3y^2 = x+3y$

Question

I am calculating the curve that is formed by the graph between the function $$f(x,y)=3x^2+4xy+y^2$$ and the plane $$z=x+3y$$ Then I'm left with the equation $3x^2+4xy+y^2=x+3y$ that I want to parameterize but I'm stuck

Answer 1

put x=r cos A, y= r sin A. now when you put it you get a relation giving r in terms of angle A , find range of r.

Answer 2

You can write if $$xy\neq 0$$ $$3\frac{x}{y}+4+\frac{y}{x}=0$$ and now substitute $$t=\frac{x}{y}$$

Answer 3

Hint:

The equation is that of a conic and more precisely an ellipse. You can center and reduce it to the form

$$\lambda_0 x^2+\lambda_1 y^2=c$$ where the lambdas are the Eigenvalues of the matrix of quadratic coefficients. You will find that the Eigenvalues are $1,5$ and Eigenvectors are $(1,1)$ and $(1,-1)$ so that the centered equation reads

$$5(\cos t+\sin t)^2+(\cos t-\sin t)^2=c.$$

Hence the requested parametric equation reads

$$x=\sqrt5(\cos t+\sin t)+a,\\y=\cos t-\sin t+b$$ where the constants $a,b$ that can be found by plugging in the initial equation.

Answer 4

As you and maverick said, taking $$x=r\cos(\theta)$$ $$y=r\sin(\theta)$$ and knowing $x^2+y^2=r^2$ you can give a parametrization of this curve.

$$3r^2\cos^2(\theta)+4r^2\cos(\theta)\sin(\theta)+r^2\sin^2(\theta)=r\cos(\theta)+3r\sin(\theta)$$

$$ r^2(1+2\cos^2(\theta)+4\cos(\theta)\sin(\theta))=r(\cos(\theta)+3\sin(\theta))$$ $$ r=\frac{\cos(\theta)+3\sin(\theta)}{1+2\cos^2(\theta)+4\cos(\theta)\sin(\theta)}$$

Hence you can parametrize your curve as:

$$ r(\theta)=\frac{\cos(\theta)+3\sin(\theta)}{1+2\cos^2(\theta)+4\cos(\theta)\sin(\theta)}$$

Remark: I do not know if this parametrization will be useful in your case, but it is a parametrization. Now, you need to define the range of $r$ related to the range of $\theta$ you have.