Family of lines `x sec^(2) theta + y tan^(2)theta -2=0` for different real `theta`, is

Options

A. Not concurrent
B. Concurrent at `(1,1)`
C. Concurrent at `(2,-2)`
D. Concurrent at `(-2,2)`

Answer

To solve the problem of finding the family of lines represented by the equation \( x \sec^2 \theta + y \tan^2 \theta - 2 = 0 \) for different real values of \( \theta \), we can follow these steps:

### Step-by-Step Solution:

1. **Rewrite the Given Equation**:
The equation given is:
\[ x \sec^2 \theta + y \tan^2 \theta - 2 = 0 \]

2. **Use the Identity for Secant**:
We know that:
\[ \sec^2 \theta = 1 + \tan^2 \theta \]
Substitute this into the equation:
\[ x(1 + \tan^2 \theta) + y \tan^2 \theta - 2 = 0 \]
This simplifies to:
\[ x + x \tan^2 \theta + y \tan^2 \theta - 2 = 0 \]

3. **Group the Terms**:
Rearranging gives:
\[ x - 2 + (x + y) \tan^2 \theta = 0 \]

4. **Identify the Family of Lines**:
This can be expressed in the form of a family of lines:
\[ L_1: x - 2 = 0 \quad \text{and} \quad L_2: x + y = 0 \]
Here, \( L_1 \) and \( L_2 \) represent two lines.

5. **Find the Point of Concurrency**:
To find the point of concurrency of these two lines, we can solve them simultaneously:
- From \( L_1 \): \( x = 2 \)
- Substitute \( x = 2 \) into \( L_2 \):
\[ 2 + y = 0 \implies y = -2 \]
Thus, the point of concurrency is:
\[ (2, -2) \]

### Conclusion:
The family of lines described by the equation \( x \sec^2 \theta + y \tan^2 \theta - 2 = 0 \) for different real \( \theta \) converges at the point \( (2, -2) \).