Parameterisation Of The Intersection Between $z=x^2+y^2+1$ And $z=2x+2y+3$

Question

Find a parameterisation of the curve of the intersection between the ellipsoid $z=x^2+y^2+1$ and the plane $z=2x+2y+3$

We have

\begin{align}x^2+y^2+1=2x+2y+3&\iff x^2-2x+y^2-2y=2\\&\iff(x-1)^2-1+(y-1)^2-1=2\\&\iff(x-1)^2+(y-1)^2=4\end{align}

So it is a circle located at $(1,1)$ of radius $2$

Can we set $x-1=2 \cos t, y-1=2 \sin t$ and say that the parameterisation of the curve is:

$$\gamma(t)=(2\cos t,2 \sin t)?$$

Answer 1

First of all, the equation $z=x^2+y^2+1$ does not describe an ellipsoid. It describes an elliptic paraboloid.

On the other hand, your $\gamma$ describes a curve in $\mathbb{R}^2$ and therefore it cannot possibly be an answer to your problem. The set of those $(x,y,z)\in\mathbb{R}^3$ such that $(x-1)^2+(y-1)^2=4$ is a cylinder, not a circle. An answer would be$$\gamma(t)=\bigl(2\cos(t)+1,2\sin(t)+1,4\cos(t)+4\sin(t)+7\bigr).$$

Answer 2

The parametrization, since it is a curve in $\mathbb{R^3}$, should be

$$\gamma(t)=[x(t),y(t),z(t)]=[2\cos t+1,2 \sin t+1,2(2\cos t+1)+2(2 \sin t+1)+3]$$

Answer 3

No, $$\gamma(t)=(2\cos t,2 \sin t)$$ does not work because

• the $z$ is missing,

• you dropped the terms $-1$.

Fixing:

$$x=2\cos x+1, \\y=2\sin x+1, \\z=x^2+y^2+1=4(\cos x+\sin x)+7=2x+2y+3.$$