If a cannon ball is fired at a target, where the line $L$ joining the point of projection to the target makes an angle $\alpha$ to the horizontal, and the target is at height $h$ above the point of projection.
Show that if the cannon ball is fired at an angle $\alpha+\beta$ to the horizontal and it hits the target in a direction perpendicular to the line $L$, then
$$\tan \alpha \tan \beta= \frac{1}{2}$$
Confused as how to tackle the question.
\begin{align*} x &= ut\cos \theta \\ y &= ut\sin \theta-\frac{gt^2}{2} \\ y &= x\tan \theta-\frac{gx^2}{2u^2\cos^2 \theta} \tag{1} \\ \frac{\dot{y}}{\dot{x}} &= \frac{u\sin \theta-g t}{u\cos \theta} \\ \frac{dy}{dx} &= \tan \theta-\frac{gx}{u^2\cos^2 \theta} \tag{2} \end{align*}
$2\times (1)-x\times (2)$, $$ 2y-x\frac{dy}{dx} = 2x\tan \theta \\$$
Now \begin{align*} x &= h\cot \alpha \\ y &= h \\ \theta &= \alpha+\beta \\ \frac{dy}{dx} &= -\cot \alpha \\ \end{align*}
Therefore \begin{align*} 2h-(h\cot \alpha)(-\cot \alpha) &= 2(h\cot \alpha)\tan (\alpha+\beta) \\ 2\tan \alpha+\cot \alpha &= \frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta} \\ 2\tan \alpha+\cot \alpha-2\tan^2 \alpha \tan \beta-\tan \beta &= \tan \alpha+\tan \beta \\ \tan \alpha+\cot \alpha &= 2\tan \beta+2\tan^2 \alpha \tan \beta \\ (1+\tan^2 \alpha)\cot \alpha &= 2(1+\tan^2 \alpha)\tan \beta \\ \tan \alpha \tan \beta &= \frac{1}{2} \end{align*}
hint...set up axes parallel to and perpendicular to the slope and write down the equations of motion. Use the condition that $\dot{x}=0$ when $y=0$