Question

The surfaces intersect in a space curve C. Determine the projection of C onto an xy plane
5x^2 +4x^2+(z-2)^2=2
5x^2+4y^2=z^2

Answer 1

\[5x^2+4y^2+(z-2)^2=2\]\[5x^2+4y^2=z^2\]\[z^2+(z-2)^2=2\]\[2z^2-4z+4=2\]\[2(z-1)^2=0\quad,z=1\]
\[5x^2+4y^2=1\quad,\quad\frac{x^2}{(1/\sqrt{5})^2}+\frac{y^2}{(1/2)^2}=1\quad-\mbox{ellipse}\]