Question

If the chord joining two points whose eccentric angles are $\alpha$ and $\beta$, cut the major axis of an ellipse at a distance $c$ from the centre, then $\tan \frac{\alpha }{2}\tan \frac{\beta }{2}=$

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

Answer 1

The correct option is B $\frac{c-a}{c+a}$

The equation of the chord joining points whose eccentric angle are $\alpha$ and $\beta$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, is $\frac{x}{a}\cos \left(\frac{\alpha +\beta }{2}\right)+\frac{y}{b}\sin \left(\frac{\alpha +\beta }{2}\right)=\cos \left(\frac{\alpha -\beta }{2}\right)$

This will cut the major axis at the point $(c,0),$ if

$\frac{c}{a}\cos \left(\frac{\alpha +\beta }{2}\right)=\cos \left(\frac{\alpha -\beta }{2}\right)$

$=\frac{\cos \left(\frac{\alpha +\beta }{2}\right)}{\cos \left(\frac{\alpha -\beta }{2}\right)}=\frac{a}{c}$

$\Rightarrow \frac{\cos \left(\frac{\alpha +\beta }{2}\right)+\cos \left(\frac{\alpha -\beta }{2}\right)}{\cos \left(\frac{\alpha +\beta }{2}\right)-\cos \left(\frac{\alpha -\beta }{2}\right)}=\frac{a+c}{a-c}$

$\tan \frac{\alpha }{2}\tan \frac{\beta }{2}=\frac{a+c}{a-c}$