Let `P(a sectheta, 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
(A) `(a^2+b^2)/a` (B) `-((a^2+b^2)/a)` (C) `(a^2+b^2)/b` (D) `-((a^2+b^2)/b)`

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))`

Explanation

To solve the problem step by step, we will follow the instructions provided in the video transcript and derive the value of \( k \) at the intersection of the normals at points \( P \) and \( Q \) on the hyperbola.

### Step 1: Define the Points
Let \( P \) and \( Q \) be defined as follows:
- \( P(a \sec \theta, b \tan \theta) \)
- \( Q(a \sec \phi, b \tan \phi) \)

Given that \( \theta + \phi = \frac{\pi}{2} \), we can express \( \phi \) as:
\[ \phi = \frac{\pi}{2} - \theta \]

### Step 2: Write the Equations of the Normals
The equation of the normal to the hyperbola at point \( P \) is given by:
\[ a x \cos \theta + b y \cot \theta = a^2 + b^2 \]
Similarly, the equation of the normal at point \( Q \) becomes:
\[ a x \cos \phi + b y \cot \phi = a^2 + b^2 \]

### Step 3: Substitute for \( \phi \)
Substituting \( \phi = \frac{\pi}{2} - \theta \) into the equation for the normal at \( Q \):
\[ a x \sin \theta + b y \tan \theta = a^2 + b^2 \]

### Step 4: Set Up the System of Equations
Now we have two equations:
1. \( a h \cos \theta + b k \cot \theta = a^2 + b^2 \) (Equation 1)
2. \( a h \sin \theta + b k \tan \theta = a^2 + b^2 \) (Equation 2)

### Step 5: Manipulate the Equations
Multiply Equation 1 by \( \sin \theta \):
\[ a h \cos \theta \sin \theta + b k \cot \theta \sin \theta = (a^2 + b^2) \sin \theta \]
This simplifies to:
\[ a h \cos \theta \sin \theta + b k \cos \theta = (a^2 + b^2) \sin \theta \]

Now, multiply Equation 2 by \( \cos \theta \):
\[ a h \sin \theta \cos \theta + b k \tan \theta \cos \theta = (a^2 + b^2) \cos \theta \]
This simplifies to:
\[ a h \sin \theta \cos \theta + b k \sin \theta = (a^2 + b^2) \cos \theta \]

### Step 6: Subtract the Two Equations
Now, subtract the modified Equation 2 from the modified Equation 1:
\[ (b k \cos \theta - b k \sin \theta) = (a^2 + b^2) (\sin \theta - \cos \theta) \]
This gives:
\[ b k (\cos \theta - \sin \theta) = (a^2 + b^2)(\sin \theta - \cos \theta) \]

### Step 7: Solve for \( k \)
Rearranging gives:
\[ b k = - (a^2 + b^2) \implies k = -\frac{a^2 + b^2}{b} \]

### Conclusion
Thus, the value of \( k \) is:
\[ k = -\frac{a^2 + b^2}{b} \]

### Final Answer
The correct option is (D) \( -\frac{(a^2 + b^2)}{b} \).