Question

Try to sketch by hand the curve of intersection of the parabolic cylinder y = x2 and the top half of the ellipsoid x2 + 7y2 + 7z2 = 49. Then find parametric equations for this curve.

Answer 1

Plug $y=x^2$ into the equation of the ellipsoid:

$x^2+7(x^2)^2+7z^2=49$

Complete the square:

$7x^4+x^2=7\left(x^4+\dfrac{x^2}7+\dfrac1{14^2}-\dfrac1{14^2}\right)=7\left(x^2+\dfrac1{14}\right)^2+\dfrac1{28}$

Then the intersection is such that

$7\left(x^2+\dfrac1{14}\right)^2+7z^2=\dfrac{1371}{28}$

$\left(x^2+\dfrac1{14}\right)^2+z^2=\dfrac{1371}{196}$

which resembles the equation of a circle, and suggests a parameterization is polar-like coordinates. Let

$x(t)^2+\dfrac1{14}=\sqrt{\dfrac{1371}{196}}\cos t\implies x(t)=\pm\sqrt{\sqrt{\dfrac{1371}{196}}\cos t-\dfrac1{14}}$

$y(t)=x(t)^2=\sqrt{\dfrac{1371}{196}}\cos t-\dfrac1{14}$

$z=\sqrt{\dfrac{1371}{196}}\sin t$

(Attached is a plot of the two surfaces and the intersection; red for the positive root $x(t)$, blue for the negative)