Posted by giules on .
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,
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,
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,
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? 
maths 
Rleigh,
Using these two points (0,2) & (8,7) can y find a & b using ghe formula y=aĆ·2(e^bx+e^bx)