Value of $4\sin^2(\alpha)$ If the point on $y=x\tan(\alpha)-\frac{ax^2}{2u^2\cos^2(\alpha)}$ ,$\alpha>0$ where the tangent is parallel to $y=x$ has an ordinate $\frac{u^2}{4a}$ then value of $4\sin^2(\alpha)$ is?. If we substitute $y$ in the given equation we get a quadratic in $x$ whose roots are $-\tan(\alpha)\pm\sqrt{\tan^2(\alpha)-1}$  . Also $\frac{dy}{dx}=1$ but after doing this I don't get an equation in $\alpha$ as there is an $u$ making things difficult . I don't know how to proceed. Note that answer is an integer from $[0-9]$. Any hints ?
 A: You must also assume $u \ne 0$ and $\cos(\alpha)\ne 0$. The slope of the tangent at any point is given by the derivative with respect to $x$, which will be a linear function and therefore will equal 1 at exactly one value of $x$.
Compute the derivative with respect to $x$ and set that equal to 1. Solve for $x$ (expressed as a function of the parameters $u$ and $\alpha$). If I did this right, you should get $x = \dfrac{\sin(\alpha)\cos(\alpha)}{\alpha} u$.
Then plug this back into the equation to compute the value of $y$ at this $x$ (that is the "ordinate" of the point where the tangent has slope 1). Set that equal to $\dfrac{u^2}{4\alpha}$; you will see that a lot of simplifications take place, in particular $u$ is eliminated from the equation. This will lead you to the final answer.
A: Let $P \equiv  \left(x,\frac{u^2}{4a}\right) $
Then by $$y=x\tan(\alpha)-\frac{ax^2}{2u^2\cos^2(\alpha)}$$
we get $$\frac{u^2}{4a}=x\tan(\alpha)-\frac{ax^2}{2u^2\cos^2(\alpha)}$$
$$\frac{u^2}{4}=ax\tan(\alpha)-\frac{(ax)^2}{2u^2\cos^2(\alpha)}\rightarrow (1)$$
And $$\frac{dy}{dx}=\tan(\alpha)-\frac{ax}{u^2\cos^2(\alpha)}$$
$$1=\tan(\alpha)-\frac{ax}{u^2\cos^2(\alpha)}$$
$$(\tan\alpha-1)(u^2\cos^2\alpha)=ax \rightarrow (2)$$
Thus by $(1)$ & $(2)$ , 
$$\frac{u^2}{4}=(\tan\alpha-1)(u^2\cos^2\alpha)\tan\alpha-\frac{(\tan\alpha-1)^2(u^2\cos^2\alpha)^2}{2u^2\cos^2\alpha}$$
$$\frac{u^2}{4}=(\tan\alpha-1)(u^2\cos^2\alpha)\tan(\alpha)-\frac{(\tan\alpha-1)^2(u^2\cos^2\alpha)}{2}$$
$$\frac{\sec^2 \alpha}{4}=(\tan\alpha-1)\tan\alpha-\frac{(\tan\alpha-1)^2}{2}$$
$$\frac{\sec^2 \alpha}{4}=\tan^2\alpha-\tan\alpha-\frac{\tan^2\alpha}{2}+\tan\alpha -\frac{1}{2}$$
$$\frac{\sec^2 \alpha}{4}=\frac{\tan^2\alpha}{2}-\frac{1}{2}$$
$$\sec^2 \alpha=2\tan^2\alpha-2$$
$$1+\tan^2 \alpha=2\tan^2\alpha-2$$
$$\tan^2\alpha =3$$
Thus $$4\sin ^2 \alpha =3$$
A: Your equation is the same as the equation of the trajectory of a projectile, with $a$ instead of $g$.
Treating it as a mechanics problem, we can parametrize the curve with
$$x=ut\cos\alpha$$ $$y=ut\sin\alpha-\frac 12 at^2$$
So we have$$\dot{x}=u\cos\alpha$$ 
$$\dot{y}=u\sin\alpha-at$$
When $\frac{dy}{dx}=1$ we can set $\dot{x}=\dot{y}\implies t=\frac ua(\sin\alpha-\cos\alpha)$
We can substitute the value of $t$ in the $y$ equation.
We then have $$\frac{u^2}{4a}=\frac{u^2}{a}(\sin^2\alpha-\sin\alpha\cos\alpha)-\frac{u^2}{a}(1-2\sin\alpha\cos\alpha)$$
This reduces to $$\sin^2\alpha=\frac 34$$
