the parabola P : y^{2} = 4ax, a > 0
We assume that the reader is amiliar with the graph of
the curve y = kx^{2}, with k > 0. This is shown in the top
sketch.
If we reflect this graph in the
line y = x, then the transformed curve will have equation
x = ky^{2}. For k = 1/4a, this is the graph of P. This is shown in the
second sketch.
We observe that the yaxis is the tangent to P at the
origin  the vertex of P.
Apart from the identity e, the only obvious symmetry
is h, reflection in the xaxis. The existence of this
symmetry is apparent from the algebra, since, if (x,y)εP,
then y = 4ax^{2}, so y = 4a(x)^{2}, i.e. (x,y)εP.
For the moment, we will take it as intuitive that there
are no other symmetries, so that the symmetry group
of P is {e,h}.
For P, the reflection symmetry is reflection in the xaxis.
This is the axis of the parabola.

