the relation of the bezier Curve and ellipse?

I want to draw a oval in html5 canvas,and i found a good method for it in stackoverflow.but I have another quesition.

function drawEllipse(ctx, x, y, w, h) {
  var kappa = 0.5522848;
      ox = (w / 2) * kappa, // control point offset horizontal
      oy = (h / 2) * kappa, // control point offset vertical
      xe = x + w,           // x-end
      ye = y + h,           // y-end
      xm = x + w / 2,       // x-middle
      ym = y + h / 2;       // y-middle

  ctx.beginPath();
  ctx.moveTo(x, ym);
  ctx.bezierCurveTo(x, ym - oy, xm - ox, y, xm, y);
  ctx.bezierCurveTo(xm + ox, y, xe, ym - oy, xe, ym);
  ctx.bezierCurveTo(xe, ym + oy, xm + ox, ye, xm, ye);
  ctx.bezierCurveTo(xm - ox, ye, x, ym + oy, x, ym);
  ctx.closePath();
  ctx.stroke();
}

the method in the above link has use bezierCurveTo to draw a ellipse,but it has draw bezierCurveTo 4 times. but I think just 2 bezierCurveTo can draw a ellipse.like this:

but i'm weak in Mathematics,could someone tell me the relationship of "the control point" and "the oval point" please? or we must draw four bezier Curve to draw a oval?

thanks everybody

My background isn't in mathematics so if I'm wrong I'm sure someone will correct me, but from my understanding we can draw a pretty good approximation of an ellipse with just two cubic bezier curves but the coordinates will be a little tricky.

To answer your question about the relation between the oval point and the control points I think it best be answered by watching this video from the point I've selected if you're familiar with interpolation or from the beginning if you are not. Don't worry it is short.

One problem we're probably going to run into is that when we start from the top and do a bezierCurveTo the bottom of the ellipse with the corners of the rectangle (of the same width and height) as the control points, the ellipses width is going to be smaller than the rectangle. .75 times the size we want. So we can just scale the control points accordingly.

Our control point's x would be adjusted like so (assuming width is the width of the ellipse and we're dividing by two to get its offset from the origin)

var cpx = (width / .75) / 2;

Put together a visualization where you can play with the width and height and see the drawn ellipse.

The red ellipse is how we wanted it to be drawn, with the inner one how it would be drawn if we didnt reposition the control points. The lines illustrate De Casteljau's algorithm that was shown in the video.

Here's a screenshot of the visualization