Prove Evolute of Rectangular Hyperbola: x=a\cosh\theta, y=a\sinh\theta

[John O' Meara]

Prove that the evolute of the rectangular hyperbola x=a\cosh\theta, y=a\sinh\theta is x^{\frac{2}{3}} - y^{\frac{2}{3}}= (2a)^{\frac{2}{3}}\\.
Let (X,Y) be a pair of coordinates on the center of curvature (the evolute) of the hyperbola. X =x -\frac{y^{'}(1 + (y^')^2)}{y^{''}} \\ and Y = y-\frac{(1 + (y^')^2)}{y^{''}} \\, where y^{'}=\frac{dy}{dx} \\.
Now \frac{dx}{d\theta} = a sinh\theta, \frac{dy}{d\theta}=a cosh\theta \\; therefore \frac{dy}{dx} = \coth\theta \\;
hence \frac{d^2y}{dx^2} = \frac{d \coth\theta}{d\theta}\frac{d\theta}{dx} =-\cosech^2\theta \frac{d\theta}{dx}\\.
Therefore y^{''} =-\frac{cosech^2 \theta}{a\sinh\theta}\\.
Therefore X = a\cosh\theta + a\cosh\theta(\sinh^2\theta + \cosh^2\theta)\\ and Y = a\sinh\theta + a\sinh\theta(\sinh^2\theta + \cosh^2\theta) \\.
Thest are the parametric equations of the evolute. How do I get the cartesian form of the equations for the evolute. I mean how do I eliminate \theta. Thanks for the help.

 

[John O' Meara]

Therefore I get x^{\frac{1}{3}} - y^{\frac{1}{3}} = (2a)^{\frac{1}{3}}((\cosh^3\theta)^{\frac{1}{3}} \\ - ( \sinh^{\frac{1}{3}}\theta + \sinh^3\theta)^{\frac{1}{3}}) \\ which gives x^{\frac{2}{3}} - y^{\frac{2}{3}} = (2a)^{\frac{2}{3}}( 1 - \sinh^{\frac{2}{3}}\theta)\\ Can anyone see where I went wrong? Thanks for the help.

 

[SanjeevGupta]

John

There is a typo in the formula for Y above. It should be Y=y + etc. Then you get the required result.

Regards

 

[John O' Meara]

Thanks for the reply.