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 sum of the squares of the lengths of the semi-conjugate diameters \( CP \) and \( CD \) of the ellipse given by the equation:

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

### Step 1: Identify the Points on the Ellipse
Let \( P \) and \( D \) be points on the ellipse corresponding to the semi-conjugate diameters. We can express these points in terms of a parameter \( \theta \):

- Point \( P \) can be represented as \( (a \cos \theta, b \sin \theta) \).
- Point \( D \) can be represented as \( (a \cos(90^\circ + \theta), b \sin(90^\circ + \theta)) \).

Using the trigonometric identities:
- \( \cos(90^\circ + \theta) = -\sin \theta \)
- \( \sin(90^\circ + \theta) = \cos \theta \)

Thus, point \( D \) can be expressed as:
\[ D = (-a \sin \theta, b \cos \theta) \]

### Step 2: Calculate \( CP^2 \)
The distance \( CP \) from the center \( C(0, 0) \) to point \( P \) is calculated as follows:

\[ CP^2 = (a \cos \theta - 0)^2 + (b \sin \theta - 0)^2 \]
\[ CP^2 = a^2 \cos^2 \theta + b^2 \sin^2 \theta \]

### Step 3: Calculate \( CD^2 \)
Similarly, the distance \( CD \) from the center \( C(0, 0) \) to point \( D \) is:

\[ CD^2 = (-a \sin \theta - 0)^2 + (b \cos \theta - 0)^2 \]
\[ CD^2 = a^2 \sin^2 \theta + b^2 \cos^2 \theta \]

### Step 4: Sum \( CP^2 + CD^2 \)
Now, we can find the sum of the squares of the distances:

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

Combining the terms:

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

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:

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