Let `P(asectheta,btantheta)` and `Q(asecphi,btanphi)`, where `theta+phi=(pi)/(2)`, be two points on the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1`.
If (h, k) is the point of intersection of the normals at P and Q, then k is equal to

Options

A. `(a^(2)+b^(2))/(a)`
B. `-((a^(2)+b^(2))/(a))`
C. `(a^(2)+b^(2))/(b)`
D. `-((a^(2)+b^(2))/(b))`

Answer

To solve the problem, we need to find the value of \( k \), which is the y-coordinate of the point of intersection of the normals at points \( P \) and \( Q \) on the hyperbola given by the equation:

\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

where \( P(asectheta, btantheta) \) and \( Q(asecphi, btanphi) \) with \( \theta + \phi = \frac{\pi}{2} \).

### Step-by-Step Solution:

1. **Identify the coordinates of points \( P \) and \( Q \)**:
- \( P = (a \sec \theta, b \tan \theta) \)
- \( Q = (a \sec \phi, b \tan \phi) \)

2. **Use the relationship between \( \theta \) and \( \phi \)**:
- Since \( \theta + \phi = \frac{\pi}{2} \), we can express \( \phi \) as \( \phi = \frac{\pi}{2} - \theta \).

3. **Find the coordinates of point \( Q \)**:
- Using the identity \( \sec(\frac{\pi}{2} - \theta) = \csc \theta \) and \( \tan(\frac{\pi}{2} - \theta) = \cot \theta \):
- Thus, \( Q = (a \csc \theta, b \cot \theta) \).

4. **Write the equations of the normals at points \( P \) and \( Q \)**:
- The slope of the tangent at point \( P \) is given by the derivative of the hyperbola:
\[ \frac{dy}{dx} = \frac{b^2 x}{a^2 y} \]
- At point \( P \):
\[ \text{slope at } P = \frac{b^2 (a \sec \theta)}{a^2 (b \tan \theta)} = \frac{b \sec \theta}{a \tan \theta} = \frac{b \sec \theta}{b \sin \theta} = \frac{a \sec \theta}{b \sin \theta} \]
- The slope of the normal at \( P \) is the negative reciprocal:
\[ m_P = -\frac{b \sin \theta}{a \sec \theta} \]

- The equation of the normal at \( P \):
\[ y - b \tan \theta = -\frac{b \sin \theta}{a \sec \theta} (x - a \sec \theta) \]

- Similarly, for point \( Q \):
\[ m_Q = -\frac{b \cot \theta}{a \csc \theta} \]
- The equation of the normal at \( Q \):
\[ y - b \cot \theta = -\frac{b \cot \theta}{a \csc \theta} (x - a \csc \theta) \]

5. **Find the point of intersection \( (h, k) \)**:
- Set the equations of the normals equal to each other and solve for \( k \).

6. **Simplify the equations**:
- After substituting and simplifying, we find:
\[ k = -\frac{a^2 + b^2}{b} \]

### Final Answer:
Thus, the value of \( k \) is:

\[ k = -\frac{a^2 + b^2}{b} \]