If the straight line `xcosalpha+ysinalpha=p` touches the curve `(x^2)/(a^2)-(y^2)/(b^2)=1` , then `p^2dot`

Options

A. `a^(2)cos^(2)alpha-b^(2)sin^(2)alpha=p^(2)`
B. `a^(2)cos^(2)alpha-b^(2)sin^(2)alpha=p`
C. `a^(2)cos^(2)alpha+b^(2)sin^(2)alpha=p^(2)`
D. `a^(2)cos^(2)alpha+b^(2)sin^(2)alpha=p`

Explanation

To solve the problem, we need to determine the value of \( p^2 \) given that the line \( x \cos \alpha + y \sin \alpha = p \) is tangent to the hyperbola defined by \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).

### Step-by-Step Solution:

1. **Rewrite the Line Equation**:
Start with the line equation:
\[ x \cos \alpha + y \sin \alpha = p \]
Rearranging gives:
\[ y \sin \alpha = p - x \cos \alpha \]
Dividing through by \( \sin \alpha \):
\[ y = \frac{p}{\sin \alpha} - \frac{x \cos \alpha}{\sin \alpha} \]
This can be expressed in slope-intercept form:
\[ y = -\frac{\cos \alpha}{\sin \alpha} x + \frac{p}{\sin \alpha} \]

2. **Identify the Slope**:
The slope \( m \) of the line is:
\[ m = -\frac{\cos \alpha}{\sin \alpha} = -\cot \alpha \]

3. **Condition for Tangency**:
For the line to be tangent to the hyperbola, the condition we need to satisfy is:
\[ c^2 = a^2 m^2 - b^2 \]
where \( c = \frac{p}{\sin \alpha} \) and \( m = -\cot \alpha \).

4. **Substituting Values**:
Substitute \( c \) and \( m \) into the tangency condition:
\[ \left(\frac{p}{\sin \alpha}\right)^2 = a^2 \left(-\cot \alpha\right)^2 - b^2 \]
Simplifying \( m^2 \):
\[ m^2 = \cot^2 \alpha = \frac{\cos^2 \alpha}{\sin^2 \alpha} \]
Thus, the equation becomes:
\[ \frac{p^2}{\sin^2 \alpha} = a^2 \frac{\cos^2 \alpha}{\sin^2 \alpha} - b^2 \]

5. **Clearing the Denominator**:
Multiply through by \( \sin^2 \alpha \):
\[ p^2 = a^2 \cos^2 \alpha - b^2 \sin^2 \alpha \]

6. **Final Result**:
Therefore, the value of \( p^2 \) is:
\[ p^2 = a^2 \cos^2 \alpha - b^2 \sin^2 \alpha \]

### Summary:
The final answer is:
\[ p^2 = a^2 \cos^2 \alpha - b^2 \sin^2 \alpha \]