<s>[INST] 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 ?
 [/INST] 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 $$4\sin ^2 \alpha =3$$
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