Question

lf the chord joining the points, whose eccentric angles are $\alpha$ and $\beta$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, meets the major axis at a distance $d$ from the centre, then $\tan \frac{\beta }{2}\tan \frac{\alpha }{2}=$

• A. $\frac{d+2a}{d-2a}$
• B. $\frac{d-2a}{d+2a}$
• C. $\frac{d-a}{d+a}$
• D. $\frac{d+a}{d-a}$

Answer 1

Answer: (C)

$\frac{d-a}{d+a}$

The points on the given ellipse with eccentric angles $\alpha$ and $\beta$ has coordinates $(a\cos \alpha ,b\sin \alpha )$ and $(a\cos \beta ,b\sin \beta )$ respectively.
Similarly, the point where it meets major axis has coordinates $(d,0)$.
Applying condition of collinearity of these points, we get
$\left|\begin{array}{ccc}a\cos \beta & b\sin \beta & 1\\ a\cos \alpha & b\sin \alpha & 1\\ d& 0& 1\end{array}\right|=0$
$\Rightarrow db(\sin \beta -\sin \alpha )+ab(\sin \alpha \cos \beta -\cos \alpha \sin \beta )=0$
$\Rightarrow db\left(2\cos \frac{\beta +\alpha }{2}\sin \frac{\beta -\alpha }{2}\right)+2ab\sin \frac{\alpha -\beta }{2}\cos \frac{\alpha -\beta }{2}=0\Rightarrow d\cos \frac{\beta +\alpha }{2}-a\cos \frac{\alpha -\beta }{2}=0\Rightarrow (d-a)\left(\cos \frac{\alpha }{2}\cos \frac{\beta }{2}\right)-(d+a)\left(\sin \frac{\alpha }{2}\sin \frac{\beta }{2}\right)=0$
.
Dividing by $\cos \frac{\alpha }{2}\cos \frac{\beta }{2}$, we get
$(d+a)\left(\tan \frac{\alpha }{2}\tan \frac{\beta }{2}\right)=d-a\Rightarrow \tan \frac{\alpha }{2}\tan \frac{\beta }{2}=\frac{d-a}{d+a}$