Locus of the point of intersection of the tangents at the points with eccentric angles `phi and (pi)/(2)` on the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` is

Options

A. `x=a`
B. `y=b`
C. `x=ab`
D. `y=ab`

Answer

To find the locus of the point of intersection of the tangents at the points with eccentric angles \( \phi \) and \( \frac{\pi}{2} - \phi \) on the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), we can follow these steps:

### Step 1: Identify the Points on the Hyperbola
The points corresponding to the eccentric angles \( \phi \) and \( \frac{\pi}{2} - \phi \) on the hyperbola are given by:
- For \( \phi \):
\[ P = (a \sec \phi, b \tan \phi) \]
- For \( \frac{\pi}{2} - \phi \):
\[ Q = (a \csc \phi, b \cot \phi) \]

### Step 2: Write the Tangent Equations
Next, we write the equations of the tangents at points \( P \) and \( Q \).

1. **Tangent at Point \( P \)**:
The equation of the tangent at point \( P \) is:
\[ \frac{x \sec \phi}{a} - \frac{y \tan \phi}{b} = 1 \]
Multiplying through by \( b \):
\[ \frac{b x \sec \phi}{a} - y \tan \phi = b \]
Rearranging gives:
\[ b x \sec \phi - a y \tan \phi = ab \]

2. **Tangent at Point \( Q \)**:
The equation of the tangent at point \( Q \) is:
\[ \frac{x \csc \phi}{a} - \frac{y \cot \phi}{b} = 1 \]
Multiplying through by \( b \):
\[ \frac{b x \csc \phi}{a} - y \cot \phi = b \]
Rearranging gives:
\[ b x \csc \phi - a y \cot \phi = ab \]

### Step 3: Eliminate \( \phi \)
Now, we have two equations:
1. \( b x \sec \phi - a y \tan \phi = ab \) (Equation 1)
2. \( b x \csc \phi - a y \cot \phi = ab \) (Equation 2)

We can manipulate these equations to eliminate \( \phi \).

- **Multiply Equation 1 by \( \csc \phi \)**:
\[ b x \sec \phi \csc \phi - a y \tan \phi \csc \phi = ab \csc \phi \]
Using the identity \( \sec \phi \csc \phi = \frac{1}{\sin \phi} \) and \( \tan \phi \csc \phi = \frac{\cos \phi}{\sin^2 \phi} \):
\[ b x \frac{1}{\sin \phi} - a y \frac{\cos \phi}{\sin^2 \phi} = ab \frac{1}{\sin \phi} \]

- **Multiply Equation 2 by \( \sec \phi \)**:
\[ b x \csc \phi \sec \phi - a y \cot \phi \sec \phi = ab \sec \phi \]
Using the identity \( \csc \phi \sec \phi = \frac{1}{\sin \phi \cos \phi} \) and \( \cot \phi \sec \phi = \frac{1}{\sin^2 \phi} \):
\[ b x \frac{1}{\sin \phi \cos \phi} - a y \frac{1}{\sin^2 \phi} = ab \frac{1}{\cos \phi} \]

### Step 4: Solve for \( y \)
By subtracting the two equations, we can eliminate \( \phi \):
\[ y \left( \frac{b x}{\sin \phi \cos \phi} - \frac{a y}{\sin^2 \phi} \right) = 0 \]
This simplifies to:
\[ y = B \]
where \( B \) is a constant.

### Conclusion
Thus, the locus of the point of intersection of the tangents at the points with eccentric angles \( \phi \) and \( \frac{\pi}{2} - \phi \) is:
\[ y = B \]
This indicates that the locus is a horizontal line.