Intersections of quadratic planes as elliptic curves An elliptic curve defined over a field $k$ is a smooth projective curve of genus $1$, plus a $k$-rational point. Every elliptic curve can be written in a Weierstrass form, i.e. as a plane cubic curve of the form $y^2 + a_1 x y + a_3 y = x^3 + a_2 x^2 + a_4 x + a_6$.
Now let $A$ and $B$ be two symmetric $4 \times 4$ matrices. Consider the curve $C$ in $\mathbb{P}^3$ defined by: $^tx A x = 0$, $^tx B x = 0$. If the homogeneous polynomial $\det(\lambda A + \mu B) \in k[\lambda, \mu]$ does not have multiple fators, then the curve $C$ is of genus $1$. Thus if there is a rational point on $C$, then one can define an elliptic curve.

My main question is: how can one deduce a Weierstrass form for such a curve?
By "how", I mean an algorithm which accepts $A$ and $B$ as inputs and gives the $a_i$'s as outputs.
For other models such as cubic plane curves, or hyperelliptic curves of degree $4$, I am aware of such algorithms. But I cannot find reference for this intersection model.

A secondary question is: is it possible to do "computations" directly on the curve $C$?
For example, if an elliptic curve is given as a cubic plane curve, i.e. by a homogeneous polynomial of degree $3$: $F(X, Y, Z) = 0$, then the addition law can be described easily by intersecting projective lines with the curve: if $O$ is the zero point and $P$, $Q$ are two points on the curve, then one can find the sum $P + Q$ as follows: intersect the line $PQ$ with the curve, call the third intersection $R$, then intersect the line $OR$ with the curve, and the third intersection is just $P + Q$.
Is there a similar procedure in the model of intersection of quadric surfaces?
 A: Regarding your main question, this is done in Cassels, Lectures on elliptic curves, $\S$ 8 (iv) p. 36. We may assume that the common rational point of the quadrics is $(X:Y:Z:T)=(0:0:0:1)$. Then the quadrics have the shape
\begin{align*}
Q_1 & = TL + R\\
Q_2 & = TM + S
\end{align*}
where $L,M$ (resp. $R,S$) are linear (resp. quadratic) in $X,Y,Z$. The polynomials $L,M$ are necessarily linearly independent, and eliminating $T$ gives the cubic equation
\begin{equation*}
P(X,Y,Z) = LS-RM=0.
\end{equation*}
So we are reduced to put a cubic into canonical form, which you know how to do.
Regarding your second question, it is indeed possible to work out the group law directly on the intersection $C$ of two quadric surfaces. This is explained in Husemöller, Elliptic curves, Introduction, $\S$ 8, p. 20. The idea is as follows: given two points $P,Q$ on $C$, consider the plane containing $O,P,Q$, and intersect it with $C$. By Bézout's theorem there will be 4 points of intersection, $O$, $P$, $Q$ and a fourth point $-(P+Q)$. The map $P \to -P$ is defined similarly.
