Two tangents are drawn from a point on hyperbola `x^(2)-y^(2)=5` to the ellipse `(x^(2))/(9)+(y^(2))/(4)=1`. If they make angle `alpha and beta` with x-axis, then

Options

A. `alpha-beta=pm(pi)/(2)`
B. `alpha+beta=(pi)/(2)`
C. `alpha+beta=pi`
D. `alpha+beta=0`

Explanation

To solve the problem, we need to analyze the tangents drawn from a point on the hyperbola \(x^2 - y^2 = 5\) to the ellipse \(\frac{x^2}{9} + \frac{y^2}{4} = 1\). We will find the angles \(\alpha\) and \(\beta\) that these tangents make with the x-axis.

### Step-by-Step Solution:

1. **Identify the equations of the hyperbola and the ellipse**:
- The hyperbola is given by \(x^2 - y^2 = 5\).
- The ellipse is given by \(\frac{x^2}{9} + \frac{y^2}{4} = 1\).

2. **Rewrite the equations in standard form**:
- For the hyperbola, we can rewrite it as:
\[ \frac{x^2}{5} - \frac{y^2}{5} = 1 \]
Here, \(a^2 = 5\) and \(b^2 = 5\), so \(a = b = \sqrt{5}\).
- For the ellipse, we have \(a^2 = 9\) and \(b^2 = 4\), so \(a = 3\) and \(b = 2\).

3. **Find the general point on the hyperbola**:
- A point on the hyperbola can be represented in parametric form as:
\[ (x, y) = (\sqrt{5} \sec \theta, \sqrt{5} \tan \theta) \]

4. **Equation of the tangent to the ellipse**:
- The equation of the tangent to the ellipse at a slope \(m\) is given by:
\[ y = mx + \sqrt{9m^2 + 4} \]

5. **Substituting the point on the hyperbola into the tangent equation**:
- The tangent line must pass through the point \((\sqrt{5} \sec \theta, \sqrt{5} \tan \theta)\):
\[ \sqrt{5} \tan \theta = m(\sqrt{5} \sec \theta) + \sqrt{9m^2 + 4} \]

6. **Rearranging the equation**:
- Rearranging gives:
\[ \sqrt{5} \tan \theta - m \sqrt{5} \sec \theta = \sqrt{9m^2 + 4} \]
- Squaring both sides results in:
\[ 5 \tan^2 \theta - 2m \cdot 5 + 5m^2 \sec^2 \theta = 9m^2 + 4 \]

7. **Simplifying the equation**:
- This leads to a quadratic equation in \(m\):
\[ (5 \sec^2 \theta - 9)m^2 - 10m \tan \theta + (5 \tan^2 \theta - 4) = 0 \]

8. **Using properties of roots**:
- Let \(m_1\) and \(m_2\) be the slopes of the tangents. The product of the roots \(m_1 m_2\) is given by:
\[ m_1 m_2 = \frac{5 \tan^2 \theta - 4}{5 \sec^2 \theta - 9} \]
- Since the tangents make angles \(\alpha\) and \(\beta\) with the x-axis, we have:
\[ \tan \alpha \tan \beta = m_1 m_2 = 1 \]

9. **Finding the sum of angles**:
- Using the identity for the tangent of the sum of angles:
\[ \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \]
- Since \(\tan \alpha \tan \beta = 1\), we have:
\[ \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{0} \]
- This implies that \(\alpha + \beta = \frac{\pi}{2}\).

### Conclusion:
The angles \(\alpha\) and \(\beta\) made by the tangents with the x-axis satisfy:
\[ \alpha + \beta = \frac{\pi}{2} \]