Write the equation
[tex]y\text{ = tan}\theta\text{ . x - }\frac{g}{2u^2.cos^2\theta}.x^2[/tex]1)
[tex]\begin{gathered} \text{Solve for }\theta \\ \sec ^2\theta\text{ = }\frac{1}{\cos^2\theta} \\ 1+tan^2\theta=sec^2\theta \\ y\text{ = xtan}\theta\text{ - }\frac{g\sec^2\theta}{2u^2}.x^2 \\ y\text{ = xtan}\theta\text{ - }\frac{gx^2}{2u^2}\sec e^2\theta \\ y\text{ = xtan}\theta\text{ - }\frac{gx^2}{2u^2}(1+tan^2\theta) \\ y\text{ = xtan}\theta\text{ - }\frac{gx^2}{2u^2\text{ }}\text{ - }\frac{gx^2}{2u^2}\tan ^2\theta \\ \frac{gx^2}{2u^2}\tan ^2\theta\text{ - xtan}\theta\text{ + y + }\frac{gx^2}{2u^2}\text{ = 0} \\ \text{Let tan}\theta\text{ = m} \\ \frac{gx^2}{2u^2}m^2\text{ - xm + y + }\frac{gx^2}{2u^2}\text{ = 0} \\ m\text{ = }\frac{x\text{ }\pm\text{ }\sqrt[]{x^2\text{ - 4(}\frac{gx^2}{2u^2})(y+\frac{gx^2}{2u^2})}}{\frac{gx}{u^2}} \\ m=sec^2\theta \\ \sec ^2\theta\text{ = }\frac{x\text{ }\pm\sqrt[]{x^2\text{ - 4(}\frac{gx^2}{2u^2})(y\text{ + }\frac{gx^2}{2u^2})}}{\frac{gx}{u^2}} \\ \\ \sec \theta\text{ = }\sqrt[]{\frac{x\text{ }\pm\sqrt[]{x^2\text{ - 4(}\frac{gx^2}{2u^2})(y+\frac{gx^2}{2u^2})}}{\frac{gx^2}{u^2}}} \\ \theta=sec^{-1}(\sqrt[]{\frac{x\pm\sqrt[]{x^2\text{ - 4(}\frac{gx^2}{2u^2})(y\text{ + }\frac{gx^2}{2u^2})}}{\frac{gx^2}{u^2}}}) \end{gathered}[/tex]2) Make x subject of the relation
[tex]\begin{gathered} y\text{ = xtan}\theta\text{ - }\frac{g}{2u^2\cos ^2\theta}x^2 \\ \frac{g}{2u^2\cos ^2\theta}x^2\text{ - xtan}\theta\text{ + y = 0} \\ x\text{ = }\frac{\tan \theta\pm\sqrt[]{\tan^2\theta-4(\frac{g}{2u^2\cos^2\theta})y}}{2(\frac{g}{2u^2\cos ^2\theta})} \\ x=\text{ }\frac{\tan \theta\text{ }\pm\text{ }\sqrt[]{\tan^2\theta\text{ - }\frac{2gy}{u^2\cos^2\theta}}}{\frac{g}{u^2\cos ^2\theta}} \end{gathered}[/tex]
