Length of the intersection of two quadric surfaces. How do I find the length of the intersection between $$ax^2+(a+1)z^2=1$$ and $$x^2+y^2+z^2=\dfrac{1}{a}\ ?$$
The idea is to make a parametrization of the curves using the variable $t$:
$$x = \sqrt{a^{-1}}\sin(t)\\
y = \pm\sqrt{\dfrac{\cos\overset{2}{\cdot}(t)}{a^2+a}}\\
z = \sqrt{(a+1)^{-1}}\cos(t)$$
But I'm not sure how to proceed from here... 
Any help would be valuable!
 A: Any point $(x,y,z)$ solving the first equation satisfies
$$x={1\over\sqrt{a}}\cos t,\quad z={1\over\sqrt{a+1}}\sin t$$
for some $t\in[0,2\pi[\ $. For such a point we then obtain from the second equation
$$y^2={1\over a}-{1\over a}\cos^2 t-{1\over a+1}\sin^2 t={1\over a(a+1)}\sin^2 t\ ,$$
and this implies
$$y=\pm{1\over\sqrt{a(a+1)}}\sin t\ .$$
(This is essentially what you have obtained yourself.) It follows that the intersection of the two given quadrics consists of the two curves
$$\gamma_\pm:\quad t\mapsto{\bf r}(t):=\left({1\over\sqrt{a}}\cos t, \ \pm{1\over\sqrt{a(a+1)}}\sin t,
\ {1\over\sqrt{a+1}}\sin t\right)\qquad(0\leq t\leq2\pi)\ .$$
One computes
$$\dot{\bf r}(t)=\left(-{1\over\sqrt{a}}\sin t, \ \pm{1\over\sqrt{a(a+1)}}\cos t,
\ {1\over\sqrt{a+1}}\cos t\right)$$
and happily obtains
$$|\dot{\bf r}(t)|^2={1\over a}\sin^2 t+\left({1\over a(a+1)}+{1\over a+1}\right)\cos^2 t={1\over a}\qquad(0\leq t\leq2\pi)\ .$$
It follows that the total length $L$ of the two curves is given by 
$$L={4\pi\over\sqrt{a}}\ .$$
A: The first surface ($E$) is an elliptical cylinder with its centerline running along the $y$-axis, and the second surface ($S$) is a sphere centered at the origin.
The surface $E$ can be parameterized by:
$$
E(u,v) = \left(  \frac{1}{\sqrt a}\sin u, \;v, \;   \frac{1}{\sqrt {a+1}}\cos u   \right)
$$
If we fix $u=u_0$, we get a straight line
$$
L(v) = \left(  \frac{1}{\sqrt a}\sin u_0, \;v, \;   \frac{1}{\sqrt {a+1}}\cos u_0   \right)
$$
Where this line intersects the sphere $S$, we have
$$
\frac{1}{a}\sin^2 u_0 + v^2 + \frac{1}{a+1}\cos^2 u_0 = \frac{1}{a}
$$
There are two solutions for $v$, corresponding to points on opposite sides of the sphere. One solution is:
$$
v = \frac{1}{\sqrt{a(a+1)}}\cos u_0
$$
Substituting this value of $v$ in the equation for $L$, we get the equation of the curve of intersection $C$:
$$
C(u) = \left(  \frac{1}{\sqrt a}\sin u, \; 
\frac{1}{\sqrt{a(a+1)}}\cos u, \;   
\frac{1}{\sqrt {a+1}}\cos u   \right)
$$
which you obtained yourself. The brute force approach is now to integrate $\|C'(u)\|$. Differentiating gives:
$$
C'(u) = \left(  
\frac{1}{\sqrt a}\cos u, \; 
-\frac{1}{\sqrt{a(a+1)}}\sin u, \;   
-\frac{1}{\sqrt {a+1}}\sin u   \right)
$$
and so 
$$
\|C'(u)\|^2 = 
\frac{1}{a}\cos^2 u + 
\frac{1}{a(a+1)}\sin^2 u + 
\frac{1}{a+1}\sin^2 u = \frac{1}{a}
$$
The curve has two loops, one on either side of the sphere. You can get the arclength of one of these loops from the integral
$$
\int_0^{2\pi} \|C'(u)\|\,du
$$
The integral turns out to be easy, but actually there's an even simpler approach ... 
The intersection curve $C$ lies in the plane $z = \sqrt{a}y$, so it is a great circle of the sphere $S$. The sphere has radius  $1/\sqrt{a}$, so the arclength of $C$ must be $2\pi/\sqrt{a}$. The complete intersection also includes another circle lying in the plane $z = -\sqrt{a}y$, and the total arclength of these two circles is $4\pi/\sqrt{a}$.
