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.Locus of point C is

Options

A. `(b^(2))/(x^(2))-(a^(2))/(y^(2))=1`
B. `(b^(2))/(x^(2))+(a^(2))/(y^(2))=1`
C. `(a^(2))/(x^(2))-(b^(2))/(y^(2))=1`
D. `(a^(2))/(x^(2))+(b^(2))/(y^(2))=1`

Answer

To find the locus of point C in the given hyperbola problem, we will follow these steps:

### Step 1: Understand the Hyperbola and its Properties
The equation of the hyperbola is given by:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
where \(a\) and \(b\) are the semi-major and semi-minor axes respectively. The eccentricity \(e\) of the hyperbola is defined as:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]

### Step 2: Parametric Representation of Points on the Hyperbola
The coordinates of a point \(R\) on the hyperbola can be represented in parametric form as:
\[ R(a \sec \theta, b \tan \theta) \]
where \(\theta\) is the eccentric angle.

### Step 3: Equation of the Tangent at Point R
The equation of the tangent to the hyperbola at point \(R\) is given by:
\[ \frac{x \sec \theta}{a} - \frac{y \tan \theta}{b} = 1 \]

### Step 4: Find Intersection Points A and B
To find the points where the tangent intersects the coordinate axes:
- For point \(A\) (where \(y = 0\)):
\[ \frac{x \sec \theta}{a} = 1 \implies x = a \sec \theta \implies A(a \sec \theta, 0) \]

- For point \(B\) (where \(x = 0\)):
\[ -\frac{y \tan \theta}{b} = 1 \implies y = -b \tan \theta \implies B(0, -b \tan \theta) \]

### Step 5: Coordinates of Point C
The coordinates of point \(C\) (the fourth vertex of rectangle \(OABC\)) can be determined as:
- The x-coordinate of \(C\) is equal to the x-coordinate of \(A\) (which is \(a \sec \theta\)) and the y-coordinate of \(C\) is equal to the y-coordinate of \(B\) (which is \(-b \tan \theta\)):
\[ C(a \sec \theta, -b \tan \theta) \]

### Step 6: Expressing h and k
Let \(h\) and \(k\) be the coordinates of point \(C\):
\[ h = a \sec \theta, \quad k = -b \tan \theta \]

### Step 7: Relating sec and tan to h and k
From the expressions for \(h\) and \(k\):
\[ \sec \theta = \frac{h}{a}, \quad \tan \theta = -\frac{k}{b} \]

### Step 8: Using Trigonometric Identity
Using the identity \(\sec^2 \theta - \tan^2 \theta = 1\):
\[ \sec^2 \theta - \tan^2 \theta = \left(\frac{h}{a}\right)^2 - \left(-\frac{k}{b}\right)^2 = 1 \]
This simplifies to:
\[ \frac{h^2}{a^2} - \frac{k^2}{b^2} = 1 \]

### Step 9: Final Locus Equation
Rearranging gives us the locus of point \(C\):
\[ \frac{h^2}{a^2} - \frac{k^2}{b^2} = 1 \]
Substituting \(h\) and \(k\) with \(x\) and \(y\) respectively, we have:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

### Conclusion
The locus of point \(C\) is given by:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]