Find the parametrization of the curve resulting from intersection of a function and a curve

Question

I have the following function $f(x,y) = 2-x^2-4y^2$ and the surface $2x+4y+z-1 = 0.$ How do i go about finding the parametrization of the curve resulting from intersection of these surfaces? I see that $f(x,y)$ is the equation of an ellipsoid. I have tried to set $ f(x,y) = z$ and go from there but i cant seem to find any parameterization.

Answer 1

We have

• $z=2-x^2-4y^2$
• $z=1-2x-4y$

then

$$2-x^2-4y^2=1-2x-4y\iff x^2-2x-1+4y^2-4y=0 \\\iff(x-1)^2+(2y-1)^2=3$$

thus we can take

• $x=\sqrt 3\cos \theta +1$
• $y=\frac{\sqrt 3}2\sin \theta +\frac12$
• $z=1-2x-4y=1-2\sqrt 3\cos \theta -2-2\sqrt 3\sin \theta -2=-3-2\sqrt 3(\cos\theta+\sin \theta)$

Answer 2

Let $\alpha(t)=(x(t),y(t),z(t))$ be the parametrization of such curve. This curve lies on the paraboloid and the plane, thus the following holds:

$$\left\{\begin{array}{rcl}z&=&-x^2-4y^2\\z&=&1-2x-4y\end{array}\right..$$

This imples that $$x^2+4y^2-2x-4y=1.$$ That is equivalent to $$\frac{1}{3}(x-1)^2+\frac{4}{3}\left(y-\frac{1}{2}\right)^2=1.$$ The last equation is an ellipse, and a parametrization for it is $$\gamma(t)=\left(1+\sqrt{3}\cos t,\frac{1}{2}+\frac{\sqrt{3}}{2}\sin t \right).$$

Since $z=1-2x-4y$, a parametrization for the curve in the intersection is $$\alpha(t)=\left(1+\sqrt{3}\cos t,\frac{1}{2}+\frac{\sqrt{3}}{2}\sin t ,-3-\sqrt{3}\cos t-2\sqrt{3}\sin t\right).$$

Answer 3

You have

• $z = f(x,y)= 2- x^2-4y^2$
• $2x+4y+z-1 = 0 \Leftrightarrow z=1-2x-4y$

It follows $$1-2x-4y = 2- x^2-4y^2 \Leftrightarrow x^2-2x+4y^2-4y = 1$$ Square completion gives $$(x-1)^2-1 +(2y-1)^2-1 = 1 \Leftrightarrow (x-1)^2 +(2y-1)^2 = 3 \Leftrightarrow$$ $$ \left( \frac{x-1}{\sqrt{3}} \right)^2 + \left( \frac{2y-1}{\sqrt{3}} \right)^2 = 1$$ Now, set $$\frac{x-1}{\sqrt{3}} = \cos \phi,\; \frac{2y-1}{\sqrt{3}} = \sin \phi$$ Solve for $x$ and $y$ and plug it into one of the given equations to get a parametrization for $z$.