Question

If the chord joining two points whose eccentric angle are $\alpha and\beta$. Cuts the major axis of the ellipse$\frac{x^2}{a^2}+y^2{b}^2$ = 1 at a distance c from the centre, then $cot\frac{\alpha }{2},cot\frac{\beta }{2}$ can be equal to

• A. 1
• B. c
• C. $\frac{c+a}{c-a}$
• D. $\frac{c}{a}$

Answer 1

Answer: (C)

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

Chord joining $\alpha ,\beta$ points on ellipse:

$\frac{x}{a}\cos \left(\frac{\alpha +\beta }{2}\right)+\frac{y}{b}\sin \left(\frac{\alpha +\beta }{2}\right)=\cos \left(\frac{\alpha -\beta }{2}\right)$

It passes through $(c,0)$

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

$\Rightarrow \frac{c}{a}=\frac{\cos \frac{\alpha }{2}\cos \frac{\beta }{2}+\sin \frac{\alpha }{2}\sin \frac{\beta }{2}}{\cos \frac{\alpha }{2}\cos \frac{\beta }{2}-\sin \frac{\alpha }{2}\sin \frac{\beta }{2}}$

By components divided

$\Rightarrow \frac{c+a}{c-a}=\frac{2\cos \frac{\alpha }{2}\cos \frac{\beta }{2}}{2\sin \frac{\alpha }{2}\sin \frac{\beta }{2}}=\cot \frac{\alpha }{2}\cot \frac{\beta }{2}$

$\Rightarrow cc.$