you are trying to relate to a 3 dimensional object with a 2 dimensional equation, not impossible but a royal pain in the kazoo. I have just spent over an hour on your question and have determined that I dont have enough information to work with. To the best of my calculations the width of the pillar at the top in not 120 cm but actually 147.19114 cm. How I arrived at these numbers is quite simple.

First solve for y on your graphing calculator, : Y = + or - (4 * x^2 - 1)^(9/2)

graph the two functions in a -2.5 to 2.5 by -2.5 to 2.5 window and trace.

This step you do the hard way because you trace x not y.

Therefore y = o when x = .5 remember that is only half the diameter, double it and you get 1meter or 100cm

so when we find y = 2 half the height x = .7359557 doubled to 1.4719114 meters. Not 1.2

Is there a faster way? Probably but I havent started calculus yet.

A thought occurs, plug in your y value on the home menu and solve for x and double.

Remembering that the equation puts the center of the pillar at the origin of a Cartesian Graph plane. ( hence the top of the 4 meter pillar is at y = 2)

Now that I have thoroughly confused and intimidated you, did I shed some light.

First solve for y on your graphing calculator, : Y = + or - (4 * x^2 - 1)^(9/2)

graph the two functions in a -2.5 to 2.5 by -2.5 to 2.5 window and trace.

This step you do the hard way because you trace x not y.

Therefore y = o when x = .5 remember that is only half the diameter, double it and you get 1meter or 100cm

so when we find y = 2 half the height x = .7359557 doubled to 1.4719114 meters. Not 1.2

Is there a faster way? Probably but I havent started calculus yet.

A thought occurs, plug in your y value on the home menu and solve for x and double.

Remembering that the equation puts the center of the pillar at the origin of a Cartesian Graph plane. ( hence the top of the 4 meter pillar is at y = 2)

Now that I have thoroughly confused and intimidated you, did I shed some light.