If the line x `Cos alpha+y Sin alpha=p` touches `x^(2)/a^(2)-y^(2)/b^(2)=1` then `a^(2) Cos^(2)alpha-b^(2) Sin^(2)alpha=`

Options

A. `2p^2`
B. `p^2`
C. p
D. 2p

Answer

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

### Step-by-Step Solution:

1. **Identify the Tangent Condition**:
The line \( x \cos \alpha + y \sin \alpha = p \) touches the hyperbola. This means that at the point of tangency, the line and the hyperbola share the same slope.

2. **Equation of the Tangent to the Hyperbola**:
The equation of the tangent to the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) at the point \( (x_1, y_1) \) is given by:
\[ \frac{x x_1}{a^2} - \frac{y y_1}{b^2} = 1 \]

3. **Express the Point of Tangency**:
Since the line touches the hyperbola, we can express the coordinates of the point of tangency \( (x_1, y_1) \) in terms of \( p \):
\[ x_1 = a^2 \cos \alpha / p, \quad y_1 = b^2 \sin \alpha / p \]

4. **Substituting into the Tangent Equation**:
Substitute \( x_1 \) and \( y_1 \) into the tangent equation:
\[ \frac{x \left(\frac{a^2 \cos \alpha}{p}\right)}{a^2} - \frac{y \left(\frac{b^2 \sin \alpha}{p}\right)}{b^2} = 1 \]
Simplifying gives:
\[ \frac{x \cos \alpha}{p} - \frac{y \sin \alpha}{p} = 1 \]

5. **Rearranging the Equation**:
Multiply through by \( p \):
\[ x \cos \alpha - y \sin \alpha = p \]
This confirms that the line \( x \cos \alpha + y \sin \alpha = p \) is indeed tangent to the hyperbola.

6. **Finding the Relationship**:
From the tangency condition, we can derive:
\[ a^2 \cos^2 \alpha - b^2 \sin^2 \alpha = p^2 \]

7. **Conclusion**:
Therefore, we conclude that:
\[ a^2 \cos^2 \alpha - b^2 \sin^2 \alpha = p^2 \]

### Final Answer:
\[ a^2 \cos^2 \alpha - b^2 \sin^2 \alpha = p^2 \]