Posted by **giules** on Tuesday, May 10, 2011 at 9:23pm.

Catenary; y= a/2(e^bx +e^-bx)

goes through the origin (0,0) and has point (-9,18) & (9,18)

solve for a and b.

- maths -
**Reiny**, Tuesday, May 10, 2011 at 11:08pm
sub in (0,0)

0 = a/2(e^0 + e^)

0 = a/2(2)

a = 0

something is not right here, a catenary of the above form does not pass through (0,0)

Only if a = 0, but then the equation would collapse to

y = 0

- maths -
**giules**, Tuesday, May 10, 2011 at 11:33pm
What about if we move the turning point (0,0) to (9,0) and the other points are (0,18) and (18,18, what then?

- maths -
**giules**, Tuesday, May 10, 2011 at 11:50pm
If we use the above points,I think

y= a/2(e^bx +e^-bx)

18 = a/2 (e^b0 + e^-b0)

18 = a/2 (e^0 + e^0)

18 = a/2 (1 + 1)

18 = a/2 (2)

18 = a

To find b;

then substitute in y= a/2(e^bx +e^-bx)

at point (9,0)

0= 18/2(e^b9 +e^-9b)

0= 9(e^9b +e^-9b)

0 = 9e^9b + 9e^-9b

0 = 9e^9b + 9/e^9b (Xe^9b)

0 = 9(e^9b)^2 + 9

let e^9b=x

ie 0 = 9x^2 + 9

9x^2 = -9

x^2 = -1

x= (-1)^1/2

Still it doesn't work out

or do I differentiate d(y)/d for a gradient of 0/ Still doesn't work.

What am I missing here?

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