HyperbolaHardBloom L3
Question
If $(asec\theta,btan\theta)$ and $(asec\phi,btan\phi)$ are the ends of a focal chord of $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$, then $tan\frac{\phi}{2}tan\frac{\phi}{2}$ can be equal to :
Options
A.$\frac{e - 1}{e + 1}$
B.$\frac{1 - e}{1 + e}$
C.$\frac{1 + e}{1 - e}$
D.$\frac{e + 1}{e - 1}$
Solution
Sol. (b,c)
Equation of chord $AB$ is $\frac{x}{a}cos\frac{\alpha - \beta}{2} - \frac{y}{b}sin\left( \frac{\alpha + \beta}{2} \right) = cos\frac{\alpha + \beta}{2}$
Put ( $\pm ae,0$ ) in above equation
$\Rightarrow \frac{cos\frac{\alpha + \beta}{2}}{cos\frac{\alpha - \beta}{2}} = \frac{e}{1}$ or $\frac{- e}{1}$
Applying componendo and dividendo, $tan\frac{\alpha}{2}tan\frac{\beta}{2} = \frac{1 - e}{1 + e}$ or $\frac{1 + e}{1 - e}$
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