Question related to elliptical angles Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 $ be an ellipse and $AB$ be a chord. Elliptical angle of A is $\alpha$ and elliptical angle of B is $\beta$. AB chord cuts the major axis at a point C. Distance of C from center of ellipse is $d$. Then the value of $\tan \frac{\alpha}{2}\tan \frac{\beta}{2}$ is
$(A)\frac{d-a}{d+a}\hspace{1cm}(B)\frac{d+a}{d-a}\hspace{1cm}(C)\frac{d-2a}{d+2a}\hspace{1cm}(D)\frac{d+2a}{d-2a}\hspace{1cm}$
I dont know much about eccentric angles, so could not attempt this question. My guess is $\alpha+\beta=\frac{\pi}{2}$. I dont know this is correct or not. Please help me in solving this question.
 A: Notice, we have the coordinates of the points $A$ & $B$ as follows $$A\equiv(a\cos\alpha, b\sin \alpha)$$
$$B\equiv(a\cos\beta, b\sin \beta)$$ Hence, the equation of the chord AB
$$y-b\sin\beta=\frac{b\sin\alpha-b\sin \beta}{a\cos\alpha-a\cos \beta}(x-a\cos \beta)$$
$$y-b\sin\beta=\frac{b}{a}\frac{\sin\alpha-\sin \beta}{\cos\alpha-\cos \beta}(x-a\cos \beta)$$ Since, chord AB intersects the x-axis at the point $C(d, 0)$, where $x=d$ & $y=0$ hence setting these values in the equation of the chord AB, we get $$0-b\sin\beta=\frac{b}{a}\frac{\sin\alpha-\sin \beta}{\cos\alpha-\cos \beta}(d-a\cos \beta)$$
$$\frac{\sin\alpha-\sin \beta}{\cos\alpha-\cos \beta}=\frac{a\sin\beta}{a\cos \beta-d}$$
Now, set the values of $\sin \alpha=\frac{2\tan \frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}}$, $\sin \beta=\frac{2\tan \frac{\beta}{2}}{1+\tan^2\frac{\beta}{2}}$,  $\cos \alpha=\frac{1-\tan^2 \frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}}$, $\cos \beta=\frac{1-\tan^2 \frac{\beta}{2}}{1+\tan^2\frac{\beta}{2}}$
I hope you can take it from here.
