If the tangents drawn from a point on the hyperbola `x^(2)-y^(2)=a^(2)-b^(2)` to ellipse `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` make angle `alpha` and `beta` with the transverse axis of the hyperbola, then

Options

A. `tan alpha* tan beta=1`
B. `tan alpha+tan beta=1`
C. `tan alpha- tan beta=1`
D. `tan alpha* tan beta= -1`

Answer

To solve the problem, we need to analyze the tangents drawn from a point on the hyperbola \(x^2 - y^2 = a^2 - b^2\) to the ellipse \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) and find the relationship between the angles \(\alpha\) and \(\beta\) that these tangents make with the transverse axis of the hyperbola.

### Step-by-Step Solution:

1. **Identify the Hyperbola and Ellipse Equations**:
- The hyperbola is given by \(x^2 - y^2 = a^2 - b^2\).
- The ellipse is given by \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\).

2. **Point on the Hyperbola**:
- Let the point on the hyperbola be \((h, k)\). This means it satisfies the hyperbola equation:
\[ h^2 - k^2 = a^2 - b^2. \]

3. **Equation of the Tangent to the Ellipse**:
- The equation of the tangent to the ellipse at a point can be expressed as:
\[ y = mx + \sqrt{a^2m^2 + b^2}. \]
- Here, \(m\) is the slope of the tangent line.

4. **Substituting the Point into the Tangent Equation**:
- The tangent line can also be expressed in terms of the point \((h, k)\):
\[ k = mh + \sqrt{a^2m^2 + b^2}. \]

5. **Rearranging the Equation**:
- Rearranging gives:
\[ k - mh = \sqrt{a^2m^2 + b^2}. \]
- Squaring both sides results in:
\[ (k - mh)^2 = a^2m^2 + b^2. \]

6. **Expanding and Rearranging**:
- Expanding the left-hand side:
\[ k^2 - 2kmh + m^2h^2 = a^2m^2 + b^2. \]
- Rearranging gives:
\[ (h^2 - a^2)m^2 - 2km + (b^2 - k^2) = 0. \]

7. **Finding the Slopes**:
- The above equation is a quadratic in \(m\). Let the slopes of the tangents be \(m_1\) and \(m_2\).
- The product of the slopes \(m_1m_2\) can be found using the quadratic formula:
\[ m_1m_2 = \frac{b^2 - k^2}{h^2 - a^2}. \]

8. **Relating Slopes to Angles**:
- Since \(m_1 = \tan(\alpha)\) and \(m_2 = \tan(\beta)\), we have:
\[ \tan(\alpha) \tan(\beta) = \frac{b^2 - k^2}{h^2 - a^2}. \]

9. **Using the Hyperbola Condition**:
- From the hyperbola condition \(h^2 - k^2 = a^2 - b^2\), we can substitute:
\[ \tan(\alpha) \tan(\beta) = 1. \]

### Conclusion:
Thus, we conclude that:
\[ \tan(\alpha) \tan(\beta) = 1. \]