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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.

Ali
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diya
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1 Answers1

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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.

Harish Chandra Rajpoot
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  • ,Sir I tried to solve it further but could not get $\tan\frac{\alpha}{2}\tan\frac{\beta}{2}$ in any of the options given. – diya Aug 13 '15 at 04:35
  • OK, what else did you get? Anyway, whatsoever you got will be correct if calculations were done correctly. No worry, the options may have some error. The main thing is to follow the correct method & calculations. – Harish Chandra Rajpoot Aug 13 '15 at 04:50