If CP and CD are semi-conjugate diameters of the ellipse `x^(2)/a^(2) + y^(2)/b^(2) = 1`, then `CP^(2) + CD^(2) =`

Options

A. `a + b`
B. `a^(2) + b^(2)`
C. `a^(2) - b^(2)`
D. `sqrt(a^(2) + b^(2))`

Explanation

To solve the problem, we need to find the expression for \( CP^2 + CD^2 \) where \( CP \) and \( CD \) are semi-conjugate diameters of the ellipse given by the equation:

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

### Step 1: Define the coordinates of points P and D
Let the coordinates of point \( P \) on the ellipse be given by:

\[ P(a \cos \theta, b \sin \theta) \]

Since \( CP \) and \( CD \) are semi-conjugate diameters, the coordinates of point \( D \) can be expressed as:

\[ D(a \cos(\theta + \frac{\pi}{2}), b \sin(\theta + \frac{\pi}{2})) \]

Using the angle addition formulas, we can simplify the coordinates of \( D \):

\[ D(a \cos(\theta + \frac{\pi}{2}), b \sin(\theta + \frac{\pi}{2})) = D(-a \sin \theta, b \cos \theta) \]

### Step 2: Calculate \( CP^2 \)
The length \( CP \) can be calculated as follows:

\[ CP^2 = (x_P - x_C)^2 + (y_P - y_C)^2 \]

Assuming \( C \) is the center of the ellipse at the origin (0, 0), we have:

\[ CP^2 = (a \cos \theta)^2 + (b \sin \theta)^2 = a^2 \cos^2 \theta + b^2 \sin^2 \theta \]

### Step 3: Calculate \( CD^2 \)
Now we calculate \( CD^2 \):

\[ CD^2 = (x_D - x_C)^2 + (y_D - y_C)^2 \]

Again, assuming \( C \) is at the origin:

\[ CD^2 = (-a \sin \theta)^2 + (b \cos \theta)^2 = a^2 \sin^2 \theta + b^2 \cos^2 \theta \]

### Step 4: Combine \( CP^2 \) and \( CD^2 \)
Now, we add \( CP^2 \) and \( CD^2 \):

\[ CP^2 + CD^2 = (a^2 \cos^2 \theta + b^2 \sin^2 \theta) + (a^2 \sin^2 \theta + b^2 \cos^2 \theta) \]

Rearranging the terms, we get:

\[ CP^2 + CD^2 = a^2 (\cos^2 \theta + \sin^2 \theta) + b^2 (\sin^2 \theta + \cos^2 \theta) \]

Using the Pythagorean identity \( \cos^2 \theta + \sin^2 \theta = 1 \):

\[ CP^2 + CD^2 = a^2 \cdot 1 + b^2 \cdot 1 = a^2 + b^2 \]

### Final Result
Thus, we conclude that:

\[ CP^2 + CD^2 = a^2 + b^2 \]