Let P any point on ellipse `3x^(2)+4y^(2)=12` . If S and S'' are its foci then find the the locus of the centroid of trianle PSS''

Explanation

To find the locus of the centroid of triangle PSS'', where P is any point on the ellipse given by the equation \(3x^2 + 4y^2 = 12\), we will follow these steps:

### Step 1: Convert the ellipse equation to standard form
The given ellipse equation is:
\[ 3x^2 + 4y^2 = 12 \]
We divide the entire equation by 12 to convert it into the standard form:
\[ \frac{x^2}{4} + \frac{y^2}{3} = 1 \]
This shows that \(a^2 = 4\) and \(b^2 = 3\).

### Step 2: Identify the semi-major and semi-minor axes
From the standard form, we find:
\[ a = 2 \quad \text{and} \quad b = \sqrt{3} \]

### Step 3: Calculate the foci of the ellipse
The distance of the foci from the center is given by \(c = \sqrt{a^2 - b^2}\):
\[ c = \sqrt{4 - 3} = \sqrt{1} = 1 \]
Thus, the foci \(S\) and \(S''\) are located at:
\[ S(1, 0) \quad \text{and} \quad S''(-1, 0) \]

### Step 4: Parametrize the point P on the ellipse
A point \(P\) on the ellipse can be represented in parametric form as:
\[ P(2\cos\theta, \sqrt{3}\sin\theta) \]

### Step 5: Find the centroid of triangle PSS''
The coordinates of the centroid \(G\) of triangle \(PSS''\) can be calculated using the formula:
\[ G\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \]
Substituting the coordinates of points \(P\), \(S\), and \(S''\):
\[ G\left(\frac{2\cos\theta + 1 - 1}{3}, \frac{\sqrt{3}\sin\theta + 0 + 0}{3}\right) \]
This simplifies to:
\[ G\left(\frac{2\cos\theta}{3}, \frac{\sqrt{3}\sin\theta}{3}\right) \]

### Step 6: Express the coordinates of the centroid in terms of h and k
Let:
\[ h = \frac{2\cos\theta}{3} \quad \text{and} \quad k = \frac{\sqrt{3}\sin\theta}{3} \]

### Step 7: Relate h and k using trigonometric identities
From the expressions for \(h\) and \(k\), we can express \(\cos\theta\) and \(\sin\theta\):
\[ \cos\theta = \frac{3h}{2} \quad \text{and} \quad \sin\theta = \frac{3k}{\sqrt{3}} = \sqrt{3}k \]

### Step 8: Use the Pythagorean identity
Using the identity \(\cos^2\theta + \sin^2\theta = 1\):
\[ \left(\frac{3h}{2}\right)^2 + (\sqrt{3}k)^2 = 1 \]
This simplifies to:
\[ \frac{9h^2}{4} + 3k^2 = 1 \]

### Step 9: Rewrite the equation in standard form
To express this in terms of \(x\) and \(y\), we can substitute \(h = x\) and \(k = y\):
\[ \frac{9x^2}{4} + 3y^2 = 1 \]

### Final Result
The locus of the centroid of triangle \(PSS''\) is given by:
\[ \frac{9x^2}{4} + 3y^2 = 1 \]