Let `P(theta_1) and Q(theta_2)` are the extremities of any focal chord of the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` whose eccentricity is e. Let `theta` be the angle between its asymptotes. Tangents are drawn to the hyperbola at some arbitrary points R. These tangent meet the coordinate axes at the points A and B respectively. The rectangle OABC (O being the origin) is completedm, then
Q. If `cos^(2)((theta_1+theta_2)/(2))=lambdacos^(2)((theta_1-theta_2)/(2))`, then `lambda` is equal to

Options

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

Answer

To solve the problem, we start with the hyperbola given by the equation:

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

### Step 1: Identify Points on the Hyperbola
Let \( P(\theta_1) \) and \( Q(\theta_2) \) be the extremities of a focal chord of the hyperbola. In polar coordinates, these points can be represented as:
- \( P = (a \sec \theta_1, b \tan \theta_1) \)
- \( Q = (a \sec \theta_2, b \tan \theta_2) \)

### Step 2: Equation of the Chord PQ
The equation of the chord \( PQ \) can be derived using the two points \( P \) and \( Q \). The equation of the chord joining these two points is given by:

\[ \frac{x}{a \cos \left(\frac{\theta_1 - \theta_2}{2}\right)} - \frac{y}{b \sin \left(\frac{\theta_1 - \theta_2}{2}\right)} = \cos \left(\frac{\theta_1 + \theta_2}{2}\right) \]

### Step 3: Substitute Point A
Since the chord passes through the focus \( (ae, 0) \), we substitute \( x = ae \) and \( y = 0 \) into the chord equation:

\[ \frac{ae}{a \cos \left(\frac{\theta_1 - \theta_2}{2}\right)} - 0 = \cos \left(\frac{\theta_1 + \theta_2}{2}\right) \]

This simplifies to:

\[ e = \cos \left(\frac{\theta_1 + \theta_2}{2}\right) \cos \left(\frac{\theta_1 - \theta_2}{2}\right) \]

### Step 4: Square Both Sides
Now, squaring both sides gives us:

\[ e^2 = \cos^2 \left(\frac{\theta_1 + \theta_2}{2}\right) \cos^2 \left(\frac{\theta_1 - \theta_2}{2}\right) \]

### Step 5: Compare with Given Equation
We are given that:

\[ \cos^2 \left(\frac{\theta_1 + \theta_2}{2}\right) = \lambda \cos^2 \left(\frac{\theta_1 - \theta_2}{2}\right) \]

From our earlier result, we can equate:

\[ e^2 = \lambda \]

### Step 6: Express \( e^2 \) in Terms of \( a \) and \( b \)
The eccentricity \( e \) of the hyperbola is given by:

\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]

Thus,

\[ e^2 = 1 + \frac{b^2}{a^2} \]

### Step 7: Final Expression for \( \lambda \)
Substituting back, we get:

\[ \lambda = 1 + \frac{b^2}{a^2} \]

This can be rewritten as:

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

### Conclusion
The value of \( \lambda \) is:

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