*Parabolic co-ordinates \((\xi, \eta)\) in a plane are defined by \(\xi=r+x, \eta=\) \(r-x\). Find \(x\) and \(y\) in terms of \(\xi\) and \(\eta\). Show that the kinetic energy of a particle of mass \(m\) is $$ T=\frac{m}{8}(\xi+\eta)\left(\frac{\dot{\xi}^{2}}{\xi}+\frac{\dot{\eta}^{2}}{\eta}\right) $$ Hence find the equations of motion.

Given the relationship between the parabolic coordinates \(\xi\) and \(\eta\) and the standard Cartesian coordinates \(x\) and \(r\): $$\xi = r + x$$ $$\eta = r - x$$ By solving these equations simultaneously, we can find expressions for \(x\) and \(y\) in terms of the parabolic coordinates. Add both equations to eliminate \(r\): $$(\xi + \eta) = 2x$$ Then, divide both sides by \(2\) to find \(x\): $$x=\frac{\xi+\eta}{2}$$ Subtract \(\eta\) from \(\xi\), to eliminate \(x\): $$(\xi - \eta) = 2r$$ Then, divide both sides by \(2\) to find \(r\): $$r=\frac{\xi-\eta}{2}$$ Now we have expressions for \(x\) and \(y\) in terms of the parabolic coordinates: $$x=\frac{\xi+\eta}{2}$$ $$r=\frac{\xi-\eta}{2}$$

Now, we will find the kinetic energy T in terms of the parabolic coordinates. The kinetic energy of a particle can be expressed as: $$T=\frac{1}{2}m\left(\dot{x}^{2} + \dot{r}^{2}\right)$$ First, take the time derivatives (\(\dot{}\)) of \(x\) and \(r\) with respect to time. $$\dot{x} = \frac{\dot{\xi} + \dot{\eta}}{2}$$ $$\dot{r} = \frac{\dot{\xi} - \dot{\eta}}{2}$$ Square these expressions of \(\dot{x}\) and \(\dot{r}\). $$\dot{x}^2 = \frac{\dot{\xi}^2 + 2\dot{\xi}\dot{\eta} + \dot{\eta}^2}{4}$$ $$\dot{r}^2 = \frac{\dot{\xi}^2 - 2\dot{\xi}\dot{\eta} + \dot{\eta}^2}{4}$$ Add both expressions to eliminate the term \(2\dot{\xi}\dot{\eta}\) $$\dot{x}^2 + \dot{r}^2 = \frac{\dot{\xi}^2 + \dot{\eta}^2}{2}$$ Substitute this into the expression for the kinetic energy T: $$T=\frac{1}{2}m\left(\frac{\dot{\xi}^2 + \dot{\eta}^2}{2}\right)$$ $$T=\frac{m}{4}\left(\dot{\xi}^2 + \dot{\eta}^2\right)$$ Now multiply this expression by \(\frac{\xi + \eta}{\xi + \eta}\) to obtain the given expression: $$T=\frac{m}{8}(\xi+\eta)\left(\frac{\dot{\xi}^{2}}{\xi}+\frac{\dot{\eta}^{2}}{\eta}\right)$$

The equations of motion can be found using the Euler-Lagrange equation: $$ \frac{d}{dt}\left(\frac{\partial T}{\partial \dot{q}_i}\right) - \frac{\partial T}{\partial q_i} = 0,\ \ i = 1, 2 $$ Where \(q_1 = \xi\) and \(q_2 = \eta\). Let us calculate the necessary derivatives for \(\xi\) first. $$\frac{\partial T}{\partial \dot{\xi}} = \frac{m}{8}\left(\frac{\xi+\eta}{\xi}\right)(\dot{\xi})$$ $$\frac{d}{dt}\left(\frac{\partial T}{\partial \dot{\xi}}\right) = \frac{m}{8}\left[\frac{(\xi+\eta)}{\xi}\right](\ddot{\xi})$$ $$\frac{\partial T}{\partial \xi} = \frac{m}{8}\left[\frac{\dot{\xi}^2}{\xi} - \frac{\dot{\xi}\dot{\eta}}{\xi}\right]$$ Now substitute these expressions into the Euler-Lagrange equation: $$\frac{m}{8}\left[\frac{(\xi+\eta)}{\xi}\right](\ddot{\xi})= \frac{m}{8}\left[\frac{\dot{\xi}^2}{\xi} - \frac{\dot{\xi}\dot{\eta}}{\xi}\right]$$ Now, we need to calculate the necessary derivatives for \(\eta\): $$\frac{\partial T}{\partial \dot{\eta}} = \frac{m}{8}\left(\frac{\xi+\eta}{\eta}\right)(\dot{\eta})$$ $$\frac{d}{dt}\left(\frac{\partial T}{\partial \dot{\eta}}\right) = \frac{m}{8}\left[\frac{(\xi+\eta)}{\eta}\right](\ddot{\eta})$$ $$\frac{\partial T}{\partial \eta} = \frac{m}{8}\left[\frac{\dot{\eta}^2}{\eta} - \frac{\dot{\xi}\dot{\eta}}{\eta}\right]$$ Substitute these expressions into the Euler-Lagrange equation for \(\eta\): $$\frac{m}{8}\left[\frac{(\xi+\eta)}{\eta}\right](\ddot{\eta})= \frac{m}{8}\left[\frac{\dot{\eta}^2}{\eta} - \frac{\dot{\xi}\dot{\eta}}{\eta}\right]$$ These two last equations are the equations of motion for the particle in parabolic coordinates.