Definition of the j-invariant of an elliptic curve

Actually, Ravi Vakil's notes give a great reason (modulo the strange constant out front). This explanation is given somewhere in the Foundations of Algebraic Geometry notes here http://math.stanford.edu/~vakil/preprints.html#coursenotes. I'm just going off memory, so don't attribute any errors I make to him.

Notice that once you prove that every elliptic curve has an affine model given by $y^2=x(x-1)(x-\lambda)$, then you know the $j$-invariant has to be independent of the different $\lambda$ you get by permuting.

You can just explicitly work it out that the six choices of lambda then are $\lambda, \frac{1}{\lambda}, 1-\lambda, \frac{1}{1-\lambda}, \frac{\lambda}{\lambda -1}, \frac{\lambda -1}{\lambda}$.

Some obvious first choices for an invariant with respect to all of these is to multiply them all together. You get $1$, oops, that isn't a good invariant since it not only doesn't depend on the choice of $\lambda$, but also is independent of curve. So you could also try adding them all, oops again, they come in pairs each adding to $1$, so they add to $3$.

So let's try the next best thing which is to sum the squares. If you check, this is exactly the $j$ invariant (but without the constant which is stuck in for characteristic $2$ reasons).

The $j$-invariant has the following classical interpretation.

Consider a model $E\subset{\Bbb P}^2$ of the elliptic curve (one knows that $E$ is a cubic). Let $P\in E$. There are 4 lines through $P$ that are tangent to $E$ and one can show that the set of cross-ratios $c$ of these 4 lines are independent of $P$. Then $$ j=\frac{(c^2-c+1)^3}{c^2(c-1)^2} $$ is invariant under the 24 permutations of the 4 tangents (which give up to 6 different values of the cross-ratio) and is the $j$-invariant of the elliptic curve, up to normalization.

When the curve $E$ is given in Legendre form $y^2=x(x-1)(x-\lambda)$ the formula for the $j$-invariant is obtained taking $P$ the point at infinity and the 4 tangents the line at infinity and the lines $x=0$, $x=1$ and $x=\lambda$.