If circumcentre of an equilateral triangle inscribed in `x^(2)/a^(2) + y^(2)/b^(2) = 1,` with vertices having eccentric angles `alpna, beta, gamma,` respectively is `(x_(1), y_(1))` then `sum cos alpha cos beta + sum sin alpha sin beta =`

Options

A. `(9h^(2))/a^(2)+(9k^(2))/b^(2)+3/2`
B. `9h^(2)-9k^(2)+a^(2)b^(2)`
C. `(9h^(2))/a^(2)+(9k^(2))/b^(2)+3`
D. `(9h^(2))/(2a^(2))+(9k^(2))/(2b^(2))-3/2`

Answer

To solve the problem step by step, we will follow the reasoning laid out in the video transcript and derive the required expression.

### Step 1: Understand the Problem
We have an ellipse given by the equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). An equilateral triangle is inscribed in this ellipse with vertices corresponding to eccentric angles \( \alpha, \beta, \gamma \). We need to find the value of \( \sum \cos \alpha \cos \beta + \sum \sin \alpha \sin \beta \).

### Step 2: Find Coordinates of the Vertices
The coordinates of the vertices of the triangle inscribed in the ellipse can be expressed as:
- Vertex \( P \) at \( (a \cos \alpha, b \sin \alpha) \)
- Vertex \( Q \) at \( (a \cos \beta, b \sin \beta) \)
- Vertex \( R \) at \( (a \cos \gamma, b \sin \gamma) \)

### Step 3: Calculate the Circumcenter
The circumcenter \( (x_1, y_1) \) of the triangle formed by these vertices can be calculated as:
\[ x_1 = \frac{1}{3}(a \cos \alpha + a \cos \beta + a \cos \gamma) = \frac{a}{3}(\cos \alpha + \cos \beta + \cos \gamma) \]
\[ y_1 = \frac{1}{3}(b \sin \alpha + b \sin \beta + b \sin \gamma) = \frac{b}{3}(\sin \alpha + \sin \beta + \sin \gamma) \]

### Step 4: Formulate the Equations
From the coordinates of the circumcenter, we can write two equations:
1. \( 3x_1 = a(\cos \alpha + \cos \beta + \cos \gamma) \)
2. \( 3y_1 = b(\sin \alpha + \sin \beta + \sin \gamma) \)

### Step 5: Square and Add the Equations
Now, we square both equations and add them:
\[ (3x_1)^2 = a^2(\cos \alpha + \cos \beta + \cos \gamma)^2 \]
\[ (3y_1)^2 = b^2(\sin \alpha + \sin \beta + \sin \gamma)^2 \]
Adding these gives:
\[ 9(x_1^2 + y_1^2) = a^2(\cos \alpha + \cos \beta + \cos \gamma)^2 + b^2(\sin \alpha + \sin \beta + \sin \gamma)^2 \]

### Step 6: Expand the Right Side
Using the identity \( (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc) \):
\[ 9(x_1^2 + y_1^2) = a^2(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + 2(\cos \alpha \cos \beta + \cos \beta \cos \gamma + \cos \gamma \cos \alpha)) + b^2(\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma + 2(\sin \alpha \sin \beta + \sin \beta \sin \gamma + \sin \gamma \sin \alpha)) \]

### Step 7: Substitute and Rearrange
Rearranging the terms, we can isolate the terms involving \( \sum \cos \alpha \cos \beta \) and \( \sum \sin \alpha \sin \beta \):
\[ 2 \sum (\cos \alpha \cos \beta + \sin \alpha \sin \beta) = \frac{9}{2} \left( \frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} \right) - \frac{3}{2} \]

### Step 8: Final Expression
Thus, we can express the desired sum as:
\[ \sum \cos \alpha \cos \beta + \sum \sin \alpha \sin \beta = \frac{9}{2} \left( \frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} \right) - \frac{3}{2} \]