PQ and QR are two focal chords of an ellipse and the eccentric angles of P,Q,R are `2alpha, 2beta, 2 gamma`, respectively then `tan beta gamma` is equal to

Options

A. `cot alpha`
B. `cot^(2)alpha`
C. `2 cot alpha`
D. None of these

Explanation

To solve the problem, we need to find the value of \( \tan \beta \tan \gamma \) given that \( P, Q, R \) are points on the ellipse with eccentric angles \( 2\alpha, 2\beta, 2\gamma \) respectively.

### Step-by-Step Solution:

1. **Equation of the Ellipse**:
The standard form of the ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
where \( a \) and \( b \) are the semi-major and semi-minor axes respectively.

2. **Coordinates of Points**:
The coordinates of points \( P, Q, R \) on the ellipse can be expressed using their eccentric angles:
- For point \( P \) (at \( 2\alpha \)):
\[ P = (a \cos 2\alpha, b \sin 2\alpha) \]
- For point \( Q \) (at \( 2\beta \)):
\[ Q = (a \cos 2\beta, b \sin 2\beta) \]
- For point \( R \) (at \( 2\gamma \)):
\[ R = (a \cos 2\gamma, b \sin 2\gamma) \]

3. **Equation of Chord PQ**:
The equation of the chord \( PQ \) can be derived as follows:
\[ \frac{x}{a \cos(\alpha + \beta)} + \frac{y}{b \sin(\alpha + \beta)} = \cos(\alpha - \beta) \]

4. **Focus of the Ellipse**:
The focus of the ellipse is located at \( (ae, 0) \) where \( e \) is the eccentricity given by \( e = \sqrt{1 - \frac{b^2}{a^2}} \).

5. **Using the Focal Property**:
Since the chord \( PQ \) passes through the focus, we can set up the equation:
\[ \frac{ae}{a \cos(\alpha + \beta)} + \frac{0}{b \sin(\alpha + \beta)} = \cos(\alpha - \beta) \]
This simplifies to:
\[ e = \cos(\alpha - \beta) \cdot \cos(\alpha + \beta) \]

6. **Setting Up Another Equation for Chord QR**:
Similarly, for the chord \( QR \):
\[ e = \cos(\beta - \gamma) \cdot \cos(\beta + \gamma) \]

7. **Equating the Two Expressions for e**:
From the two expressions for \( e \), we have:
\[ \cos(\alpha - \beta) \cdot \cos(\alpha + \beta) = -\cos(\alpha - \gamma) \cdot \cos(\alpha + \gamma) \]

8. **Applying Componendo and Dividendo**:
By applying the method of componendo and dividendo, we can derive:
\[ \frac{\cos(\alpha + \beta) + \cos(\alpha - \beta)}{\cos(\alpha + \beta) - \cos(\alpha - \beta)} = \frac{\cos(\alpha + \gamma) - \cos(\alpha - \gamma)}{\cos(\alpha + \gamma) + \cos(\alpha - \gamma)} \]

9. **Solving for \( \tan \beta \tan \gamma \)**:
After simplification, we find:
\[ \tan \beta \tan \gamma = \cot^2 \alpha \]

### Final Result:
Thus, we conclude that:
\[ \tan \beta \tan \gamma = \cot^2 \alpha \]