To find the locus of the centroid of triangle \( PSS'' \) where \( P \) is any point on the ellipse \( 3x^2 + 4y^2 = 12 \) and \( S \) and \( S'' \) are its foci, we will follow these steps:
### Step 1: Rewrite the equation of the ellipse in standard form
The given equation of the ellipse is:
\[
3x^2 + 4y^2 = 12
\]
Dividing the entire equation by 12, we get:
\[
\frac{x^2}{4} + \frac{y^2}{3} = 1
\]
This shows that \( a^2 = 4 \) and \( b^2 = 3 \).
### Step 2: Identify the values of \( a \) and \( b \)
From \( a^2 = 4 \) and \( b^2 = 3 \), we find:
\[
a = 2 \quad \text{and} \quad b = \sqrt{3}
\]
### Step 3: Find the foci of the ellipse
The foci \( S \) and \( S'' \) of the ellipse are given by the coordinates \( (\pm c, 0) \), where \( c = \sqrt{a^2 - b^2} \).
Calculating \( c \):
\[
c = \sqrt{4 - 3} = \sqrt{1} = 1
\]
Thus, the foci are:
\[
S(1, 0) \quad \text{and} \quad S''(-1, 0)
\]
### Step 4: Parametrize the point \( P \) on the ellipse
Using the parametric form of the ellipse, we can express the coordinates of point \( P \) as:
\[
P(2 \cos \theta, \sqrt{3} \sin \theta)
\]
### Step 5: Find the coordinates of the centroid \( G \) of triangle \( PSS'' \)
The centroid \( G \) of triangle \( PSS'' \) with vertices at \( P(2 \cos \theta, \sqrt{3} \sin \theta) \), \( S(1, 0) \), and \( S''(-1, 0) \) is given by:
\[
G\left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)
\]
Substituting the coordinates:
\[
G\left( \frac{2 \cos \theta + 1 - 1}{3}, \frac{\sqrt{3} \sin \theta + 0 + 0}{3} \right) = G\left( \frac{2 \cos \theta}{3}, \frac{\sqrt{3} \sin \theta}{3} \right)
\]
### Step 6: Express \( H \) and \( K \) in terms of \( \theta \)
Let:
\[
H = \frac{2 \cos \theta}{3}, \quad K = \frac{\sqrt{3} \sin \theta}{3}
\]
### Step 7: Eliminate \( \theta \) to find the locus
From the equations for \( H \) and \( K \):
\[
\cos \theta = \frac{3H}{2}, \quad \sin \theta = \frac{3K}{\sqrt{3}}
\]
Using the identity \( \cos^2 \theta + \sin^2 \theta = 1 \):
\[
\left(\frac{3H}{2}\right)^2 + \left(\frac{3K}{\sqrt{3}}\right)^2 = 1
\]
Simplifying gives:
\[
\frac{9H^2}{4} + \frac{9K^2}{3} = 1
\]
Multiplying through by 12 to eliminate the fractions:
\[
27H^2 + 36K^2 = 12
\]
Dividing by 12:
\[
\frac{9H^2}{4} + 3K^2 = 1
\]
### Final Result
Thus, the locus of the centroid \( G \) is:
\[
9H^2 + 12K^2 = 1
\]
In terms of \( x \) and \( y \):
\[
\frac{9x^2}{4} + 3y^2 = 1
\]
