If the straight line `xcosalpha+ysinalpha=p` touches the curve `(x^2)/(a^2)+(y^2)/(b^2)=1` , then prove that `a^2cos^2alpha+b^2sin^2alpha=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 prove that \( a^2 \cos^2 \alpha + b^2 \sin^2 \alpha = p^2 \) given that the line \( x \cos \alpha + y \sin \alpha = p \) touches the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), we can follow these steps:

### Step 1: Understand the equations
The equation of the ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
The equation of the line is:
...