8 A particle is projected with speed \(49 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of elevation \(\theta\) from a point \(O\) on a horizontal plane, and moves freely under gravity. The horizontal and upward vertical displacements of the particle from \(O\) at time \(t\) seconds after projection are \(x \mathrm {~m}\) and \(y \mathrm {~m}\) respectively.

1. Express \(x\) and \(y\) in terms of \(\theta\) and \(t\), and hence show that $$y = x \tan \theta - \frac { x ^ { 2 } \left( 1 + \tan ^ { 2 } \theta \right) } { 490 } .$$
   \includegraphics[max width=\textwidth, alt={}]{35477eb8-59e0-4de6-889c-1f5841f65eec-4_627_1249_1699_447}
   The particle passes through the point where \(x = 70\) and \(y = 30\). The two possible values of \(\theta\) are \(\theta _ { 1 }\) and \(\theta _ { 2 }\), and the corresponding points where the particle returns to the plane are \(A _ { 1 }\) and \(A _ { 2 }\) respectively (see diagram).
2. Find \(\theta _ { 1 }\) and \(\theta _ { 2 }\).
3. Calculate the distance between \(A _ { 1 }\) and \(A _ { 2 }\).