If `alpha,beta` are the eccentric angles of the extremities of a focal chord of the ellipse `x^(2)/16 + y^(2)/9 = 1`, then tan `(alpha/2) tan(beta/2)`=

Options

A. `(sqrt7+4)/(sqrt7-4)`
B. `-9/23`
C. `(sqrt5-4)/(sqrt5+4)`
D. `(8sqrt7-23)/9`

Answer

To solve the problem, we need to find the value of \( \tan \left( \frac{\alpha}{2} \right) \tan \left( \frac{\beta}{2} \right) \) for the given ellipse \( \frac{x^2}{16} + \frac{y^2}{9} = 1 \).

### Step 1: Identify the parameters of the ellipse
The equation of the ellipse can be rewritten as:
\[ \frac{x^2}{4^2} + \frac{y^2}{3^2} = 1 \]
From this, we identify:
- \( a = 4 \) (semi-major axis)
- \( b = 3 \) (semi-minor axis)

### Step 2: Calculate the eccentricity \( e \)
The eccentricity \( e \) of the ellipse is given by the formula:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Substituting the values of \( a \) and \( b \):
\[ e = \sqrt{1 - \frac{3^2}{4^2}} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4} \]

### Step 3: Use the property of focal chords
For a focal chord of an ellipse, the relationship between the eccentric angles \( \alpha \) and \( \beta \) is given by:
\[ \tan \left( \frac{\alpha}{2} \right) \tan \left( \frac{\beta}{2} \right) = \frac{e - 1}{e + 1} \]

### Step 4: Substitute the value of \( e \)
Now, substituting \( e = \frac{\sqrt{7}}{4} \) into the equation:
\[ \tan \left( \frac{\alpha}{2} \right) \tan \left( \frac{\beta}{2} \right) = \frac{\frac{\sqrt{7}}{4} - 1}{\frac{\sqrt{7}}{4} + 1} \]

### Step 5: Simplify the expression
To simplify, we first find a common denominator:
\[ = \frac{\sqrt{7} - 4}{\sqrt{7} + 4} \]

### Step 6: Final answer
Thus, the value of \( \tan \left( \frac{\alpha}{2} \right) \tan \left( \frac{\beta}{2} \right) \) is:
\[ \frac{\sqrt{7} - 4}{\sqrt{7} + 4} \]