A smooth tube whose axis is horizontal has an elliptic cross-section in the form of the curve with parametric equations \[ x=a\cos\theta\qquad y=b\sin\theta \] where the \(x\)-axis is horizontal and the \(y\)-axis is vertically upwards. A particle moves freely under gravity on the inside of the tube in the plane of this cross-section. By first finding \(\ddot{x}\) and \(\ddot{y},\) or otherwise, show that the acceleration along the inward normal at the point with parameter \(\theta\) is \[ \frac{ab\dot{\theta}^{2}}{\sqrt{a^{2}\sin^{2}\theta+b^{2}\cos^{2}\theta}}. \] The particle is projected along the surface in the vertical cross-section plane, with speed \(2\sqrt{bg},\) from the lowest point. Given that \(2a=3b,\) show that it will leave the surface at the point with parameter \(\theta\) where \[ 5\sin^{3}\theta+12\sin\theta-8=0. \]