Intersection of an ellipsoid and plane in parametric form I want to find the parametric equation of the ellipse in 3d space which is formed by the intersection of a known ellipsoid and a known plane. The ellipsoid has the Cartesian equation: $(x/a)^2+(y/b)^2+(z/a)^2=1$. 
While the plane has the equation: $mx+ny+kz=0$. I have substituted one equation in the other but what I get is an elliptical cylinder since I had eliminated the $z$-components. How can I get the exact equation of the ellipse of intersection in parametric form?
 A: Hint. Find vectors of the semiaxes of the ellipse, meaning find (x,y,z) which satisfies both equations and its length is longest(shortest).
$$(x/a)^2+(y/b)^2+(z/a)^2=1$$
$$mx+ny+kz=0$$
$$x^2+y^2+z^2=\min \lor \max$$
And then parametric equation will be $(x,y,z)=\vec v_1\sin \theta+\vec v_2\cos \theta$
Edit - More than a hint
As yota suggested this hint isn't too helpful since problem is still hard to solve I will try to push it till the end:
$$x^2+y^2+z^2=x^2+y^2+z^2-a^2((x/a)^2+(y/b)^2+(z/a)^2-1)=y^2(1-a^2/b^2)+a^2$$
As we can see one of the semiaxes vectors has $y$ coordinate equal to $0$, since the expression above reaches its max/min when $y=0$
We can put it in the equations to find $x$ and $z$:
$$z=-mx/k$$
$$(x/a)^2+(-mx/k/a)^2=1$$
$$x^2k^2+x^2m^2=k^2a^2$$
$$x^2=k^2a^2/(k^2+m^2)$$
So we have one semiaxis $v_1=(ka/\sqrt{k^2+m^2},0,-ma/\sqrt{k^2+m^2})$
Another should be colinear to $v_1\times(m,n,k)$ or even better $(k,0,-m)\times(m,n,k)$ which is:
$$v_2=t(mn,-m^2-k^2,kn)$$
where
$$t=1/\sqrt{(mn/a)^2+((m^2+k^2)/b)^2+(kn/a)^2}$$
