Question

If any two chords be drawn through two points on the major axis of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ equidistant from the centre. If $\alpha ,\beta ,\gamma ,\delta$ are the eccentric angles of the extremities of the chords, then the value of $\tan \frac{\alpha }{2}\tan \frac{\beta }{2}\tan \frac{\gamma }{2}\tan \frac{\delta }{2}$, is

• A. $-1$
• B. $1$
• C. $\frac{a}{b}$
• D. $0$

Answer 1

Answer: (B)

$1$

The equation of chord whose eccentric angles are $\alpha$ and $\beta$ is

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

Let it cut the positive direction of $x$ axis at a distance $c$, then

$\frac{c}{a}\cos \left(\frac{\alpha +\beta }{2}\right)+\frac{0}{b}\sin \left(\frac{\alpha +\beta }{2}\right)=\cos \left(\frac{\alpha -\beta }{2}\right)\Rightarrow \frac{c}{a}\cos \left(\frac{\alpha +\beta }{2}\right)=\cos \left(\frac{\alpha -\beta }{2}\right)\Rightarrow \frac{\cos \left(\frac{\alpha +\beta }{2}\right)}{\cos \left(\frac{\alpha -\beta }{2}\right)}=\frac{a}{c}\Rightarrow \frac{\frac{a}{c}+1}{\frac{a}{c}-1}=\frac{\{\frac{\cos \left(\frac{\alpha +\beta }{2}\right)}{\cos \left(\frac{\alpha -\beta }{2}\right)}+1\}}{\{\frac{\cos \left(\frac{\alpha +\beta }{2}\right)}{\cos \left(\frac{\alpha -\beta }{2}\right)}-1\}}\Rightarrow \frac{a+c}{a-c}=\frac{\cos \left(\frac{\alpha +\beta }{2}\right)+\cos \left(\frac{\alpha -\beta }{2}\right)}{\cos \left(\frac{\alpha +\beta }{2}\right)-\cos \left(\frac{\alpha -\beta }{2}\right)}\Rightarrow -\frac{2\cos \alpha /2\cos \beta /2}{2\sin \alpha /2\sin \beta /2}=\frac{a+c}{a-c}\Rightarrow \tan \frac{\alpha }{2}\tan \frac{\beta }{2}=\frac{c-a}{c+a}$ ......(i)

Similarly equation of chord whose eccentric angle are $\gamma$ and $\delta$ is

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

Let it cuts the axis at at distance $-c$ from the origin . then

$\tan \frac{\gamma }{2}\tan \frac{\delta }{2}=\frac{-c-a}{-c+a}=\frac{c+a}{c-a}$ .....(ii)

On multiplying (i) and (ii), we get

$\Rightarrow \tan \frac{\alpha }{2}\tan \frac{\beta }{2}\tan \frac{\gamma }{2}\tan \frac{\delta }{2}=\frac{c-a}{c+a}\times \frac{c+a}{c-a}=1$

So, option B is correct.