At the point of intersection of the rectangular hyperbola `xy=c^2` and the parabola `y^2=4ax` tangents to the rectangular hyperbola and the parabola make angles `theta` and `phi` , respectively with x-axis, then

Options

A. `theta = tan^(-1) (-2 tan phi)`
B. `theta = (1)/(2) tan^(-1)(-tan phi)`
C. `phi = tan^(-1)(-2tan theta)`
D. `phi =(1)/(2) tan^(-1)(-tan theta)`

Explanation

To solve the problem, we need to find the angles \( \theta \) and \( \phi \) made by the tangents to the rectangular hyperbola \( xy = c^2 \) and the parabola \( y^2 = 4ax \) at their point of intersection.

### Step-by-Step Solution:

1. **Equations of the Curves**:
- The equation of the rectangular hyperbola is given by:
\[ xy = c^2 \quad \text{(1)} \]
- The equation of the parabola is given by:
\[ y^2 = 4ax \quad \text{(2)} \]

2. **Finding the Point of Intersection**:
- From equation (1), we can express \( y \) in terms of \( x \):
\[ y = \frac{c^2}{x} \]
- Substitute this expression for \( y \) into equation (2):
\[ \left(\frac{c^2}{x}\right)^2 = 4ax \]
- Simplifying this gives:
\[ \frac{c^4}{x^2} = 4ax \]
- Multiplying through by \( x^2 \):
\[ c^4 = 4ax^3 \]
- Thus, we can solve for \( x \):
\[ x = \left(\frac{c^4}{4a}\right)^{\frac{1}{3}} \quad \text{(3)} \]

3. **Finding \( y \)**:
- Substitute \( x \) back into the expression for \( y \):
\[ y = \frac{c^2}{x} = \frac{c^2}{\left(\frac{c^4}{4a}\right)^{\frac{1}{3}}} = 4a^{\frac{1}{3}}c^{\frac{2}{3}} \quad \text{(4)} \]

4. **Finding the Slopes of the Tangents**:
- For the hyperbola \( xy = c^2 \):
- The slope of the tangent at the point \( (x, y) \) is given by:
\[ \text{slope} = -\frac{y}{x} = -\frac{4a^{\frac{1}{3}}c^{\frac{2}{3}}}{\left(\frac{c^4}{4a}\right)^{\frac{1}{3}}} = -\frac{4a^{\frac{1}{3}}c^{\frac{2}{3}} \cdot 4^{\frac{1}{3}}}{c^{\frac{4}{3}}} = -\frac{16a^{\frac{1}{3}}}{c^{\frac{2}{3}}} \quad \text{(5)} \]
- Therefore, \( \tan \phi = -\frac{16a^{\frac{1}{3}}}{c^{\frac{2}{3}}} \).

- For the parabola \( y^2 = 4ax \):
- The slope of the tangent at the point \( (x, y) \) is given by:
\[ \text{slope} = \frac{y}{2a} = \frac{4a^{\frac{1}{3}}c^{\frac{2}{3}}}{2a} = \frac{2c^{\frac{2}{3}}}{a^{\frac{2}{3}}} \quad \text{(6)} \]
- Therefore, \( \tan \theta = \frac{2c^{\frac{2}{3}}}{a^{\frac{2}{3}}} \).

5. **Relating \( \tan \theta \) and \( \tan \phi \)**:
- From the tangent values, we can establish the relationship:
\[ \tan \theta = -2 \tan \phi \quad \text{(7)} \]

6. **Final Values**:
- The angles \( \theta \) and \( \phi \) can be expressed as:
\[ \theta = \tan^{-1}(-2 \tan \phi) \quad \text{and} \quad \phi = \tan^{-1}\left(-\frac{1}{2} \tan \theta\right) \]