To solve the problem step by step, we will find the locus of the midpoint of the segment RN, where R trisects the line segment PQ and N is the foot of the perpendicular from R to the x-axis.
### Step 1: Identify the Points
Let \( P = (-3, 0) \) and \( Q = (0, 3t) \).
### Step 2: Find the Coordinates of Point R
Since R trisects PQ and is closer to Q, we can find the coordinates of R using the section formula. The coordinates of R can be calculated as follows:
\[
R = \left( \frac{2x_1 + x_2}{3}, \frac{2y_1 + y_2}{3} \right)
\]
Substituting \( P(-3, 0) \) and \( Q(0, 3t) \):
\[
R = \left( \frac{2(-3) + 0}{3}, \frac{2(0) + 3t}{3} \right) = \left( \frac{-6}{3}, \frac{3t}{3} \right) = (-2, t)
\]
### Step 3: Find the Equation of Line PQ
The slope of line PQ can be calculated as:
\[
\text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3t - 0}{0 - (-3)} = \frac{3t}{3} = t
\]
Using point-slope form, the equation of line PQ is:
\[
y - 0 = t(x + 3) \implies y = tx + 3t
\]
### Step 4: Find the Perpendicular Line RN
The slope of line RN, which is perpendicular to PQ, will be the negative reciprocal of the slope of PQ:
\[
\text{slope of RN} = -\frac{1}{t}
\]
Using point-slope form again, the equation of line RN passing through R(-2, t) is:
\[
y - t = -\frac{1}{t}(x + 2)
\]
Rearranging gives:
\[
y - t = -\frac{1}{t}x - \frac{2}{t} \implies y = -\frac{1}{t}x + t - \frac{2}{t}
\]
### Step 5: Find the Intersection Point N with the x-axis
To find N, set \( y = 0 \):
\[
0 = -\frac{1}{t}x + t - \frac{2}{t}
\]
Solving for x gives:
\[
\frac{1}{t}x = t - \frac{2}{t} \implies x = t^2 - 2
\]
Thus, the coordinates of N are:
\[
N = (t^2 - 2, 0)
\]
### Step 6: Find the Midpoint of RN
The midpoint M of segment RN can be calculated as:
\[
M = \left( \frac{x_R + x_N}{2}, \frac{y_R + y_N}{2} \right) = \left( \frac{-2 + (t^2 - 2)}{2}, \frac{t + 0}{2} \right)
\]
This simplifies to:
\[
M = \left( \frac{t^2 - 4}{2}, \frac{t}{2} \right)
\]
### Step 7: Eliminate t to Find the Locus
Let \( H = \frac{t^2 - 4}{2} \) and \( K = \frac{t}{2} \). From \( K = \frac{t}{2} \), we have \( t = 2K \). Substitute this into the equation for H:
\[
H = \frac{(2K)^2 - 4}{2} = \frac{4K^2 - 4}{2} = 2K^2 - 2
\]
Rearranging gives:
\[
2K^2 - H - 2 = 0 \implies K^2 = \frac{H + 2}{2}
\]
### Final Locus Equation
Thus, the locus of the midpoint M is given by:
\[
K^2 = \frac{H + 2}{2}
\]
This can be rewritten as:
\[
K^2 = \frac{1}{2}H + 1
\]
### Final Answer
The locus of the midpoint of RN is:
\[
Y^2 = \frac{1}{2}X + 1
\]
