Find the locus of the vertices of equilateral triangle circumscribing the ellipse `x^(2)/a^(2)+y^(2)/b^(2)=1`.

Answer

To find the locus of the vertices of an equilateral triangle circumscribing the ellipse given by the equation:

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

we can follow these steps:

### Step 1: Understand the properties of the triangle and ellipse
The vertices of the equilateral triangle circumscribing the ellipse will be located at a distance from the center of the ellipse. The centroid of the triangle coincides with the circumcenter due to the symmetry of the equilateral triangle.

### Step 2: Set up the coordinates
Let the coordinates of the centroid (which is also the circumcenter) of the triangle be \( G(h, k) \). For an equilateral triangle with vertices \( P, Q, R \), the coordinates can be expressed in terms of angles \( \alpha, \beta, \gamma \).

### Step 3: Express the coordinates of the centroid
The coordinates of the centroid \( G \) can be expressed as:
\[ h = \frac{1}{3}(x_1 + x_2 + x_3) \quad \text{and} \quad k = \frac{1}{3}(y_1 + y_2 + y_3) \]
where \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \) are the vertices of the triangle.

### Step 4: Use the properties of the ellipse
For the vertices of the triangle, we can express them in polar coordinates relative to the ellipse:
\[ x_i = a \cos \theta_i, \quad y_i = b \sin \theta_i \quad \text{for } i = 1, 2, 3 \]
where \( \theta_1, \theta_2, \theta_3 \) are the angles corresponding to the vertices of the triangle.

### Step 5: Calculate the coordinates of the centroid
Using the coordinates of the vertices, we can find:
\[ h = \frac{1}{3}(a \cos \theta_1 + a \cos \theta_2 + a \cos \theta_3) = \frac{a}{3}(\cos \theta_1 + \cos \theta_2 + \cos \theta_3) \]
\[ k = \frac{1}{3}(b \sin \theta_1 + b \sin \theta_2 + b \sin \theta_3) = \frac{b}{3}(\sin \theta_1 + \sin \theta_2 + \sin \theta_3) \]

### Step 6: Use the circumcenter formula
For an equilateral triangle, the circumcenter \( G(h, k) \) can also be expressed in terms of the angles:
\[ h = \frac{a^2 - b^2}{4a} \left( \cos \theta_1 + \cos \theta_2 + \cos \theta_3 \right) + \cos\left(\theta_1 + \theta_2 + \theta_3\right) \]
\[ k = \frac{b^2 - a^2}{4b} \left( \sin \theta_1 + \sin \theta_2 + \sin \theta_3 \right) - \sin\left(\theta_1 + \theta_2 + \theta_3\right) \]

### Step 7: Combine the equations
From the equations for \( h \) and \( k \), we can derive relationships between \( h \) and \( k \). By squaring and adding the equations, we can utilize the identity \( \cos^2 + \sin^2 = 1 \).

### Step 8: Derive the locus equation
After manipulating the equations, we arrive at the locus of the vertices of the triangle:
\[ \frac{h^2}{\frac{a^2 + 3b^2}{a^2(a^2 - b^2)}} + \frac{k^2}{\frac{3a^2 + b^2}{b^2(a^2 - b^2)}} = 1 \]

### Final Step: Replace \( h \) and \( k \) with \( x \) and \( y \)
Thus, the locus of the vertices of the equilateral triangle circumscribing the ellipse is given by:
\[ \frac{x^2}{\frac{a^2 + 3b^2}{a^2(a^2 - b^2)}} + \frac{y^2}{\frac{3a^2 + b^2}{b^2(a^2 - b^2)}} = 1 \]