If the line `x cos alpha + y sin alpha = p`, is tangent to the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))` = 1, then the value of `a^(2) cos^(2) alpha+b^(2) sin^(2) alpha` is

Options

A. p
B. `p^(2)`
C. `(1)/(p^(2))`
D. none of these

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 \( x \cos \alpha + y \sin \alpha = p \) is tangent to the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).

### Step-by-Step Solution:

1. **Identify the given line equation**: The line is given by \( x \cos \alpha + y \sin \alpha = p \). We can rewrite this in slope-intercept form as:
\[ y = -\frac{\cos \alpha}{\sin \alpha} x + \frac{p}{\sin \alpha} \]
Here, the slope \( m \) is \( -\frac{\cos \alpha}{\sin \alpha} \) and the y-intercept \( c \) is \( \frac{p}{\sin \alpha} \).

2. **Use the condition for tangency**: For a line \( y = mx + c \) to be tangent to the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), the condition is:
\[ c^2 = a^2 m^2 + b^2 \]
Substituting for \( m \) and \( c \):
\[ \left(\frac{p}{\sin \alpha}\right)^2 = a^2 \left(-\frac{\cos \alpha}{\sin \alpha}\right)^2 + b^2 \]

3. **Simplify the equation**:
\[ \frac{p^2}{\sin^2 \alpha} = a^2 \frac{\cos^2 \alpha}{\sin^2 \alpha} + b^2 \]
Multiply through by \( \sin^2 \alpha \):
\[ p^2 = a^2 \cos^2 \alpha + b^2 \sin^2 \alpha \]

4. **Rearranging the equation**: From the above equation, we can isolate the term we are interested in:
\[ a^2 \cos^2 \alpha + b^2 \sin^2 \alpha = p^2 \]

5. **Conclusion**: Therefore, the value of \( a^2 \cos^2 \alpha + b^2 \sin^2 \alpha \) is:
\[ \boxed{p^2} \]