how do you graph this ellipse : 4x^2 + 16^2 = 1?

Answer 1

You need a y component usually under both squares.

Answer 2

I believe you are asking for the graph of 4x^2 + 16y^2 = 1, so I will proceed with that assumption.

Ultimately, you want your equation to have the form (x-h)^2 / a^2 + (y-k)^2 / b^2 = 1. The ellipse is centered at (h, k). The shape of the ellipse is determined by a and b.

First - is the right hand side of the equation 1? If not, divide to get it equal to 1. In this case, the right side is already 1, so we move on to step two.

Second - Notice from our form that there should be no multiplication - only division. We can accomplish this using fractions: x^2 / (1/4) + y^2 / (1/16) = 1. Notice that this fits our form, with h = 0 and k = 0.

Since h = 0 and k = 0, then our ellipse is centered at the origin. Also, note that a = 1/2 (since a^2 = 1/4) and b = 1/4 (since b^2 = 1/16). Use a to determine the distance between the center and the left and right vertices. Use b to determine the distance between the top and bottom vertices.

If you would like to see the graph, go to http://www.geocities.com/louis7223/files/20061210.gif.

Answer 3

It is not an ellipse, it is a parabola. You need a y^2 component for it to be an ellipse.