If the eccentric angles of the ends of a focal chord of the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1(agtb)` are `theta_(1) and theta_(2)`, then value of `"tan"(theta_(1))/(2) "tan"(theta_(2))/(2)` equals

Options

A. `(e-1)/(e+1)`
B. `(e-1)/(e^(2)+1)`
C. `(e+1)/(e-1)`
D. `(e^(2)+1)/(e-1)`

Answer

To solve the problem, we need to find the value of \(\frac{\tan(\theta_1/2) \tan(\theta_2/2)}\) for the ends of a focal chord of the ellipse given by the equation:

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \quad (a > b) \]

### Step-by-Step Solution:

1. **Understanding the Ellipse**:
The given ellipse has its major axis along the x-axis and minor axis along the y-axis. The foci of the ellipse are located at \((\pm c, 0)\), where \(c = \sqrt{a^2 - b^2}\).

2. **Eccentric Angles**:
Let the eccentric angles of the ends of the focal chord be \(\theta_1\) and \(\theta_2\). The coordinates of the points on the ellipse corresponding to these angles are:
- For \(\theta_1\): \(P(a \cos \theta_1, b \sin \theta_1)\)
- For \(\theta_2\): \(Q(a \cos \theta_2, b \sin \theta_2)\)

3. **Focal Chord Property**:
A focal chord is a line segment that passes through one of the foci. For the ellipse, the property of the focal chord states that:
\[ \tan\left(\frac{\theta_1}{2}\right) \tan\left(\frac{\theta_2}{2}\right) = \frac{e - 1}{e + 1} \]
where \(e = \frac{c}{a} = \sqrt{1 - \frac{b^2}{a^2}}\).

4. **Using the Eccentricity**:
Substitute \(e\) into the equation:
\[ \tan\left(\frac{\theta_1}{2}\right) \tan\left(\frac{\theta_2}{2}\right) = \frac{\sqrt{1 - \frac{b^2}{a^2}} - 1}{\sqrt{1 - \frac{b^2}{a^2}} + 1} \]

5. **Final Result**:
Therefore, the value of \(\tan\left(\frac{\theta_1}{2}\right) \tan\left(\frac{\theta_2}{2}\right)\) is given by:
\[ \tan\left(\frac{\theta_1}{2}\right) \tan\left(\frac{\theta_2}{2}\right) = \frac{e - 1}{e + 1} \]

### Conclusion:
The final answer is:
\[ \frac{\tan(\theta_1/2) \tan(\theta_2/2)} = \frac{e - 1}{e + 1} \]