The area of triangle formed by the lines `x-y=0, x+y=0` and any tangent to the hyperbola `x^(2)-y^(2)=16` is equal to

Options

A. 2 sq. units
B. 4 sq. units
C. 8 sq. units
D. 16 sq. units

Answer

To find the area of the triangle formed by the lines \(x - y = 0\), \(x + y = 0\), and any tangent to the hyperbola \(x^2 - y^2 = 16\), we can follow these steps:

### Step 1: Identify the lines and their intersections
The equations of the lines are:
1. \(x - y = 0\) (or \(y = x\))
2. \(x + y = 0\) (or \(y = -x\))

These two lines intersect at the origin \((0, 0)\).

### Step 2: Find the equation of the tangent to the hyperbola
The hyperbola given is \(x^2 - y^2 = 16\). The standard form of this hyperbola is \(\frac{x^2}{16} - \frac{y^2}{16} = 1\).

Using the parametric equations for the hyperbola, we can express a point on the hyperbola as:
\[ (x_1, y_1) = (4 \sec \theta, 4 \tan \theta) \]
The equation of the tangent line to the hyperbola at this point is given by:
\[ x x_1 - y y_1 = a^2 \]
Substituting \(x_1\) and \(y_1\):
\[ x (4 \sec \theta) - y (4 \tan \theta) = 16 \]
This simplifies to:
\[ 4x \sec \theta - 4y \tan \theta = 16 \]
Dividing through by 4 gives:
\[ x \sec \theta - y \tan \theta = 4 \]

### Step 3: Find the intersection points of the tangent with the lines
1. **Intersection with \(x - y = 0\)**:
Substitute \(y = x\) into the tangent equation:
\[ x \sec \theta - x \tan \theta = 4 \]
Factoring out \(x\) gives:
\[ x (\sec \theta - \tan \theta) = 4 \implies x = \frac{4}{\sec \theta - \tan \theta} \]
Thus, the intersection point is:
\[ A\left(\frac{4}{\sec \theta - \tan \theta}, \frac{4}{\sec \theta - \tan \theta}\right) \]

2. **Intersection with \(x + y = 0\)**:
Substitute \(y = -x\) into the tangent equation:
\[ x \sec \theta + x \tan \theta = 4 \]
Factoring out \(x\) gives:
\[ x (\sec \theta + \tan \theta) = 4 \implies x = \frac{4}{\sec \theta + \tan \theta} \]
Thus, the intersection point is:
\[ B\left(\frac{4}{\sec \theta + \tan \theta}, -\frac{4}{\sec \theta + \tan \theta}\right) \]

### Step 4: Calculate the area of the triangle
The area \(A\) of triangle formed by points \(O(0, 0)\), \(A\), and \(B\) can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the coordinates of points \(A\) and \(B\):
\[ = \frac{1}{2} \left| 0\left(\frac{4}{\sec \theta - \tan \theta} - (-\frac{4}{\sec \theta + \tan \theta})\right) + \frac{4}{\sec \theta - \tan \theta}\left(-\frac{4}{\sec \theta + \tan \theta} - 0\right) + \frac{4}{\sec \theta + \tan \theta}\left(0 - \frac{4}{\sec \theta - \tan \theta}\right) \right| \]

This simplifies to:
\[ = \frac{1}{2} \left| -\frac{16}{(\sec \theta - \tan \theta)(\sec \theta + \tan \theta)} + \frac{16}{(\sec \theta + \tan \theta)(\sec \theta - \tan \theta)} \right| \]
\[ = \frac{1}{2} \left| -\frac{16}{1} \right| = 8 \]

### Final Result
The area of the triangle formed by the lines and the tangent to the hyperbola is:
\[ \text{Area} = 16 \text{ square units} \]