Intersection of sphere of radius c and a general ellipsoid

Question

Question: Given the ellipsoid $$E:\ \ \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1, \ \ 0 < a < c < b,$$

let $\pmb{\alpha}$ denotes the intersection of $E\ $ with the sphere $$S^2(c): \ \ x^2 + y^2 +z^2 =c^2.$$ Show that $\pmb{\alpha}$ is the union [of the traces] of two regular curves $\pmb{\alpha_1}$ and $\pmb{\alpha_2}$. Parametrise either curve and compute its torsion, $\tau$.

Attempts: I've not gotten past the first part. I've found that a local parametrisation for $E$ is $\mathbf{x}(u,v)=(a\cos u \sin v,\ b \sin u \sin v,\ c \ \cos v)$ which hasn't really made a difference in my confusion. In both the books I have and on related questions found on this site, the intersection seems to be given specifically for the unit sphere and I'm not sure how to extend the theory to a general sphere.
Further, any computation I do seems to return the solution $a=b=c=1$ which clearly contradicts the fact that $0 < a < c < b$.

Answer 1

Consider the coordinate system (cylindrical except that $y$ and $z$ axes are interchanged) $$ x = \rho \cos(\theta),\ y = y, z = \rho \sin(\theta)$$ and let $$u = \frac{\cos^2(\theta)}{a^2} + \frac{\sin^2(\theta)}{c^2}$$ Your ellipsoid is $$ \rho^2 u + \frac{y^2}{b^2} = 1 \tag{1}$$ while the sphere is $$\rho^2 + y^2 = c^2 \tag{2}$$ Solve for $\rho^2$ and $y^2$ in terms of $u$ (and thus in terms of $\theta$): $$ \rho^2 = \frac{b^2-c^2}{b^2u-1},\ y^2 = \frac{b^2 (c^2 u - 1)}{b^2 u - 1} $$ Note that $$ \frac{1}{b^2} \le \frac{1}{c^2} \le u \le \frac{1}{a^2} $$ so $\rho^2 > 0$ and $y^2 \ge 0$ for any $\theta$. One of your curves has $y$ the positive square root of $b^2 (c^2 u - 1)/(b^2 u - 1)$, the other has the negative square root ($\rho$ on the other hand is always positive by definition).

Answer 2

you asked this question 10 months ago, so perhaps this answer is of no use to you now, but I present my solution here:

The points $(x,y,z)$ on the ellipsoid satisfy : $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2} = 1$ and similarly The points $(x,y,z)$ on the sphere satisfy : $\frac{x^2}{c^2}+\frac{y^2}{c^2}+\frac{z^2}{c^2} = 1$.

Since $1=1$, we equate the two equations to get:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2} = \frac{x^2}{c^2}+\frac{y^2}{c^2}+\frac{z^2}{c^2}$

Cancelling the z term we get:

$\frac{x^2}{a^2}+\frac{y^2}{b^2} = \frac{x^2}{c^2}+\frac{y^2}{c^2}$

which can be re-arranged for a relation of $y$ and $x$:

$y^2 = \frac{c^2-a^2}{b^2-c^2}x^2$

The co-efficient of $x^2$ here is positive since $a<c<b$. Then taking square roots, we get 2 solutions:

$y = \pm(\frac{c^2-a^2}{b^2-c^2})^{1/2}x$

These are 2 distinct curves. Parameterising one of them shouldn't be too difficult, hope that helps :)