Tangents at any point P is drawn to hyperbola `(x^(2))/(a^(2)) - (y^(2))/(b^(2)) =1` intersects asymptotes at Q and R, if O is the centre of hyperbola then

Options

A. Area of triangle OQR is ab
B. Area of triangle OQR is 2ab
C. P is mid-point of QR
D. P trisect QR

Answer

To find the area of the triangle formed by the tangents at point P on the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) and the asymptotes, we will follow these steps:

### Step 1: Identify the Asymptotes
The asymptotes of the hyperbola are given by the equations:
\[ y = \frac{b}{a} x \quad \text{and} \quad y = -\frac{b}{a} x \]

### Step 2: General Point on the Hyperbola
A general point \( P \) on the hyperbola can be represented as:
\[ P(a \sec \theta, b \tan \theta) \]

### Step 3: Equation of the Tangent at Point P
The equation of the tangent to the hyperbola at point \( P \) is given by:
\[ \frac{x \cdot a \sec \theta}{a^2} - \frac{y \cdot b \tan \theta}{b^2} = 1 \]
This simplifies to:
\[ \frac{x \sec \theta}{a} - \frac{y \tan \theta}{b} = 1 \]

### Step 4: Intersecting the Tangent with the Asymptotes
We will find the intersection points \( Q \) and \( R \) of the tangent with the asymptotes.

#### For the first asymptote \( y = \frac{b}{a} x \):
Substituting \( y \) in the tangent equation:
\[ \frac{x \sec \theta}{a} - \frac{\frac{b}{a} x \tan \theta}{b} = 1 \]
This simplifies to:
\[ \frac{x \sec \theta}{a} - \frac{x \tan \theta}{a} = 1 \]
Factoring out \( x \):
\[ \frac{x (\sec \theta - \tan \theta)}{a} = 1 \implies x = \frac{a}{\sec \theta - \tan \theta} \]
Substituting \( x \) back to find \( y \):
\[ y = \frac{b}{a} \cdot \frac{a}{\sec \theta - \tan \theta} = \frac{b}{\sec \theta - \tan \theta} \]
Thus, the coordinates of point \( Q \) are:
\[ Q\left(\frac{a}{\sec \theta - \tan \theta}, \frac{b}{\sec \theta - \tan \theta}\right) \]

#### For the second asymptote \( y = -\frac{b}{a} x \):
Substituting \( y \) in the tangent equation:
\[ \frac{x \sec \theta}{a} + \frac{\frac{b}{a} x \tan \theta}{b} = 1 \]
This simplifies to:
\[ \frac{x \sec \theta}{a} + \frac{x \tan \theta}{a} = 1 \]
Factoring out \( x \):
\[ \frac{x (\sec \theta + \tan \theta)}{a} = 1 \implies x = \frac{a}{\sec \theta + \tan \theta} \]
Substituting \( x \) back to find \( y \):
\[ y = -\frac{b}{a} \cdot \frac{a}{\sec \theta + \tan \theta} = -\frac{b}{\sec \theta + \tan \theta} \]
Thus, the coordinates of point \( R \) are:
\[ R\left(\frac{a}{\sec \theta + \tan \theta}, -\frac{b}{\sec \theta + \tan \theta}\right) \]

### Step 5: Area of Triangle OQR
The area of triangle \( OQR \) can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Where \( O(0, 0) \), \( Q\left(\frac{a}{\sec \theta - \tan \theta}, \frac{b}{\sec \theta - \tan \theta}\right) \), and \( R\left(\frac{a}{\sec \theta + \tan \theta}, -\frac{b}{\sec \theta + \tan \theta}\right) \).

Substituting the coordinates:
\[ \text{Area} = \frac{1}{2} \left| 0 + \frac{a}{\sec \theta - \tan \theta}\left(-\frac{b}{\sec \theta + \tan \theta}\right) + \frac{a}{\sec \theta + \tan \theta}\left(\frac{b}{\sec \theta - \tan \theta}\right) \right| \]
This simplifies to:
\[ \text{Area} = \frac{1}{2} \left| \frac{ab}{(\sec \theta - \tan \theta)(\sec \theta + \tan \theta)} - \frac{ab}{(\sec \theta + \tan \theta)(\sec \theta - \tan \theta)} \right| = \frac{ab}{2} \]

### Final Result
Thus, the area of triangle \( OQR \) is:
\[ \text{Area} = ab \]