Question

Consider the surface given by

Answer 1

This is just simple. For example you have a plane of the form x=a, then you just substitute x with a, and you'll get an equation with y and z only, hence you have a 2-d trace of the intersection. It is just similar for y=b and z=c.

(1) At z=1.5, 2x^2 + 5y^2 + 1.5^2 = 4 
    2x^2 + 5y^2 = 1.75
    Now you have an ellipse in the z=1.5 plane as your trace.

(2) At x=1, 2(1)^2 + 5y^2 + z^2 = 4
    5y^2 + z^2 = 2
    Now you have an ellipse in the x=1 plane as your trace.


(3) At z=0, 2x^2 + 5y^2 + (0)^2 = 4
    2x^2 + 5y^2 = 4
    Now you have an ellipse in the z=0 plane as your trace.

(4) At y=0, 2x^2 + 5(0)^2 + z^2 = 4
    2x^2 + z^2 = 4
    Now you have an ellipse in the y=0 plane as your trace.