Question 1080821
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<pre>
One set is 

x(t) = 4*cos(t),
y(t) = 2*sin(t).


Then {{{x^2(t) + y^2(t)}}} = {{{16*cos^2(t) + 16*sin^2(t) - 16}}} = {{{16*1 -16}}} = 0.



Another set is 

x(t) = 4*sin(t),
y(t) = 2*cos(t).
</pre>

Solved.