Locus of the point of intersection of the tangents at the points with eccentric angles `phi and (pi)/(2) - phi` 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 \( \frac{\pi}{2} - \phi \) can be expressed in parametric form.

- For the angle \( \phi \):
\[ P(\phi) = (a \sec \phi, b \tan \phi) \]

- For the angle \( \frac{\pi}{2} - \phi \):
\[ Q\left(\frac{\pi}{2} - \phi\right) = (a \csc \phi, b \cot \phi) \]

### Step 2: Write the equations of the tangents
The equation of the tangent to the hyperbola at a point \( (x_0, y_0) \) is given by:
\[ \frac{xx_0}{a^2} - \frac{yy_0}{b^2} = 1 \]

- For point \( P(\phi) \):
\[ \frac{x(a \sec \phi)}{a^2} - \frac{y(b \tan \phi)}{b^2} = 1 \]
Simplifying gives:
\[ \frac{x \sec \phi}{a} - \frac{y \tan \phi}{b} = 1 \tag{1} \]

- For point \( Q\left(\frac{\pi}{2} - \phi\right) \):
\[ \frac{x(a \csc \phi)}{a^2} - \frac{y(b \cot \phi)}{b^2} = 1 \]
Simplifying gives:
\[ \frac{x \csc \phi}{a} - \frac{y \cot \phi}{b} = 1 \tag{2} \]

### Step 3: Solve the equations simultaneously
Now we have two equations:
1. \( \frac{x \sec \phi}{a} - \frac{y \tan \phi}{b} = 1 \)
2. \( \frac{x \csc \phi}{a} - \frac{y \cot \phi}{b} = 1 \)

To eliminate \( x \) and \( y \), we can manipulate these equations.

### Step 4: Multiply and rearrange
Multiply equation (1) by \( \csc \phi \) and equation (2) by \( \sec \phi \):
- From (1):
\[ \frac{x \sec \phi \csc \phi}{a} - \frac{y \tan \phi \csc \phi}{b} = \csc \phi \]

- From (2):
\[ \frac{x \sec \phi \csc \phi}{a} - \frac{y \cot \phi \sec \phi}{b} = \sec \phi \]

### Step 5: Subtract the equations
Now, subtract the two modified equations:
\[ -\frac{y \tan \phi \csc \phi}{b} + \frac{y \cot \phi \sec \phi}{b} = \csc \phi - \sec \phi \]

### Step 6: Factor out \( y \)
Factoring out \( y \):
\[ y \left(-\frac{\tan \phi \csc \phi}{b} + \frac{\cot \phi \sec \phi}{b}\right) = \csc \phi - \sec \phi \]

### Step 7: Simplify and find \( y \)
This leads to:
\[ y \left(-\frac{\sin \phi}{\cos^2 \phi} + \frac{1}{\sin^2 \phi}\right) = \csc \phi - \sec \phi \]
After simplification, we find that \( y = b \).

### Conclusion
The locus of the point of intersection of the tangents at the points with eccentric angles \( \phi \) and \( \frac{\pi}{2} - \phi \) is:
\[ y = b \]