Question

If $(asec\theta .btan\theta )$ and $(asec\Phi ,btan\Phi )$ are the ends of a focal chord of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ whose eccentricity is e,then $tan\frac{\theta }{2}\times tan\frac{\oslash }{2}$ equal to.

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Here the task will be simplified if we know the form of

a chord joining 2 points of hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.which is,

$\frac{x}{a}.cos\left[\frac{\theta -\oslash }{2}\right]-\frac{y}{b}.sin\left[\frac{\theta +\oslash }{2}\right]=cos\left[\frac{\theta +\oslash }{2}\right]$

This passes through (ae,0)

$e.cos\left[\frac{\theta -\oslash }{2}\right]-0=cos[\frac{\theta +\oslash }{2}.]$

$\frac{cos\left[\frac{\theta -\oslash }{2}\right]}{cos\left[\frac{\theta -\oslash }{2}\right]}=e$

i.e.,$\frac{e}{1}=\frac{cos\frac{\theta }{2}.cos\frac{\oslash }{2}-\frac{\theta }{2}.sin\frac{\oslash }{2}}{cos\frac{\theta }{2}.cos\frac{\oslash }{2}+\frac{\theta }{2}.sin\frac{\oslash }{2}}$

i.e.,$\frac{e+1}{e-1}=\frac{2.cos\frac{\theta }{2}.cos\frac{\oslash }{2}}{-2.sin\frac{\theta }{2}.sin\frac{\oslash }{2}}$

$\therefore tan\frac{\theta }{2}.tan\frac{\oslash }{2}=\frac{1-e}{1+e}$

Hence option (b)