Question

Consider the paraboloid z=x2+y2. the plane 4x−8y+z−10=0 cuts the paraboloid, its intersection being a curve. find "the natural" parametrization of this curve.

Answer 1

$4x-8y+z-10=0\implies4x-8y+x^2+y^2-10=0$

Complete the square to get

$x^2+4x+4-4+y^2-8y+16-16-10=0$

$(x+2)^2+(y-4)^2=30$

Now let

$x+2=\sqrt{30}\cos t$

$y-4=\sqrt{30}\sin t$

and the intersection can be expressed as a function of $t$,

$\mathbf r(t)=(\sqrt{30}\cos t-2,\sqrt{30}\sin t+4)$

with $0\le t\le2\pi$.