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`

Explanation

To prove that if the straight line \( x \cos \alpha + y \sin \alpha = p \) touches the curve \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), then \( a^2 \cos^2 \alpha + b^2 \sin^2 \alpha = p^2 \), we can follow these steps:

### Step 1: Understand the given equations
The equation of the ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
This represents an ellipse centered at the origin with semi-major axis \( b \) and semi-minor axis \( a \).

The equation of the line is:
\[ x \cos \alpha + y \sin \alpha = p \]
This represents a straight line in the Cartesian plane.

### Step 2: Write the equation of the tangent to the ellipse
The equation of the tangent to the ellipse in parametric form can be expressed as:
\[ \frac{x \cos \theta}{a} + \frac{y \sin \theta}{b} = 1 \]
where \( \theta \) is the parameter corresponding to the point of tangency on the ellipse.

### Step 3: Set the two equations equal
Since the line touches the ellipse, we can equate the two expressions:
\[ \frac{x \cos \theta}{a} + \frac{y \sin \theta}{b} = 1 \quad \text{and} \quad x \cos \alpha + y \sin \alpha = p \]

### Step 4: Compare coefficients
From the two equations, we can compare the coefficients of \( x \) and \( y \):
1. For \( x \):
\[ \frac{\cos \theta}{a} = \cos \alpha \quad \Rightarrow \quad \cos \theta = a \cos \alpha \]
2. For \( y \):
\[ \frac{\sin \theta}{b} = \sin \alpha \quad \Rightarrow \quad \sin \theta = b \sin \alpha \]
3. For the constant term:
\[ 1 = p \]

### Step 5: Square the equations
Now, we square both equations obtained from the coefficients:
1. From \( \cos \theta = a \cos \alpha \):
\[ \cos^2 \theta = a^2 \cos^2 \alpha \]
2. From \( \sin \theta = b \sin \alpha \):
\[ \sin^2 \theta = b^2 \sin^2 \alpha \]

### Step 6: Add the squared equations
Adding these two equations gives:
\[ \cos^2 \theta + \sin^2 \theta = a^2 \cos^2 \alpha + b^2 \sin^2 \alpha \]
Since \( \cos^2 \theta + \sin^2 \theta = 1 \), we have:
\[ 1 = a^2 \cos^2 \alpha + b^2 \sin^2 \alpha \]

### Step 7: Relate to \( p \)
From the constant term comparison, we know that:
\[ p^2 = 1 \]
Thus, we can conclude:
\[ a^2 \cos^2 \alpha + b^2 \sin^2 \alpha = p^2 \]

### Conclusion
We have proved that if the line touches the ellipse, then:
\[ a^2 \cos^2 \alpha + b^2 \sin^2 \alpha = p^2 \]