geometry question

Lets say we have a Parabola…X^2

and a Catenary…cosh X ((e^2-e^-2)/2)
Further, subtract 1 from the catenary so that the vertex is at the origin (0,0)

Basically what I’m saying is the parabola and the catenary are laid on top of each other on the cartesian coordinate system.

Obviously, the vertexes share the same point of zero, but what other points intersect?

I haven’t done geometry in a while, but I graphed both in Apple’s handy graphing calculator, and it appears that they don’t intersect at any points other than the vertex, assuming I graphed 'em right. But, for the “why”, you’ll have to ask someone with a better memory than I…

cosh(X) - 1 = X^2 when X = 2.98 (approximately).

The catenary = a*cosh(x/a)

And its vertex is already at x = 0.

Setting a equal to 1 you have to solve
x[sup]2[/sup] = cosh(x).

The only way I know how to solve this is numerically. It turns out they intersect at x = +/- 2.5939 and +/- 1.6213

Oops sorry you said at the origin not at x = 0.

Then the answer is as ZenBeam said x = +/- 2.983 and 0.

enolancooper, I’m sure you know what the form of cosh(x) is, so I’m not trying to inform you, but I do want to clear one thing up for anyone who doesn’t.

cosh(x) = (e[sup]x[/sup] + e[sup]-x[/sup])/2

What you have listed there is this:

(e[sup]2[/sup] - e[sup]-2[/sup])/2 = sinh(2)

And while the others here has answered the question very well, I would like to add that if you want to find the points of intersection without a graphing calculator, you can do so by finding the zero of this function:

cosh(x) - 1 - x[sup]2[/sup]

Newton’s Method is the simplest way I know. Start with some x, say 3, and keep taking the value

x - (cosh(x) - 1 - x[sup]2[/sup])/(sinh(x) - 2x)

and using that for your new value of x. The numerator is your function and the denominator is the derivative of your function. Soon you’ll get the answer. I did it, and here’s what I got:


You know you’ve got it when you get the same thing twice in a row.