Eliminating the Parameter in a Hyperbolic Curve

[Sarangalex]

Homework Statement

a) Eleminate the parameter to find a Cartesian equation of the curve.

b) Sketch the curve and indicate with an arrow the direction in which the curve is traced as the parameter increases.

Homework Equations

x = 2cosh(t) y = 5sinh(t)
cosh^2(x) - sinh^2(x) = 1

The Attempt at a Solution

cosh(t) = x/2 sinh(t) = y/5

(x^(2)/4) - (y^(2)/25) = 1

y = +/- sqrt(25/4(x^2) - 25)

This is the best guess I have here, and the graphs look similar... I just really don't know what to do from here (or even if I'm going about it the right way). Any help would be appreciated.

 

[Dick]

Homework Statement

a) Eleminate the parameter to find a Cartesian equation of the curve.

b) Sketch the curve and indicate with an arrow the direction in which the curve is traced as the parameter increases.

Homework Equations

x = 2cosh(t) y = 5sinh(t)
cosh^2(x) - sinh^2(x) = 1

The Attempt at a Solution

cosh(t) = x/2 sinh(t) = y/5

(x^(2)/4) - (y^(2)/25) = 1

y = +/- sqrt(25/4(x^2) - 25)

This is the best guess I have here, and the graphs look similar... I just really don't know what to do from here (or even if I'm going about it the right way). Any help would be appreciated.

I think you are doing fine so far. What part of this are you having a problem with?

 

[Sarangalex]

I think you are doing fine so far. What part of this are you having a problem with?

I graphed the Cartesian equation(s) I got, and the graphs didn't exactly match up.

 

[Dick]

I graphed the Cartesian equation(s) I got, and the graphs didn't exactly match up.

In what way didn't they match up? I hope in y = +/- sqrt(25/4(x^2) - 25) you meant y = +/- sqrt((25x^2)/4 - 25). You didn't use enough parentheses in the 25/4(x^2) to make it clear what you meant.

 

[Sarangalex]

In what way didn't they match up? I hope in y = +/- sqrt(25/4(x^2) - 25) you meant y = +/- sqrt((25x^2)/4 - 25). You didn't use enough parentheses in the 25/4(x^2) to make it clear what you meant.

I'm sorry; that is what I was meaning to type. Well, the parametric graph only takes up the first quadrant; however, the Cartesian graph takes up all four.

Cartesian graph:

http://www3.wolframalpha.com/Calculate/MSP/MSP419219ih3hd5eb510179000016g700d6idee9f8f?MSPStoreType=image/gif&s=4&w=200&h=204&cdf=RangeControl

 

[Dick]

I'm sorry; that is what I was meaning to type. Well, the parametric graph only takes up the first quadrant; however, the Cartesian graph takes up all four.

Cartesian graph:

http://www3.wolframalpha.com/Calculate/MSP/MSP419219ih3hd5eb510179000016g700d6idee9f8f?MSPStoreType=image/gif&s=4&w=200&h=204&cdf=RangeControl

That sort of thing would be normal. It happens when you square things. You wind up getting more solutions than you started with. If you take 0<t<infinity you are only going to get the first quadant of this graph. If you take -infinity<t<infinity then you should get both quadrants with x>0. The parametric form won't give you any of the points where x<0 since cosh(t)>0. If you really want the cartesian form to reflect that then you could solve for x in terms of y and then just take the positive square root.

 

[Sarangalex]

That sort of thing would be normal. If you take 0<t<infinity you are only going to get the first quadant of this graph. If you take -infinity<t<infinity then you should get both quadrants with x>0. The parametric form won't give you any of the points where x<0 since cosh(t)>0. If you really want the cartesian form to reflect that then you could solve for x in terms of y and then just take the positive square root.

Oh, I see. Thank you very much for your help.