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Step-by-Step Solution
Step 1: Identify Points A and B
• Let the fixed point be A with coordinates (0, 6).
• Let B be a moving point with coordinates (2t, 0).
Step 2: Find the Midpoint M of Segment AB
The midpoint M of AB is given by
$M \left(\dfrac{x_A + x_B}{2}, \dfrac{y_A + y_B}{2}\right).$
Substituting A(0, 6) and B(2t, 0):
$M \left(\dfrac{0 + 2t}{2}, \dfrac{6 + 0}{2}\right) = (t,\, 3).$
Step 3: Determine the Perpendicular Bisector of AB
1. The slope of AB is
$\text{slope of AB} = \dfrac{0 - 6}{2t - 0} = \dfrac{-6}{2t} = -\dfrac{3}{t}.$
2. The slope of the perpendicular bisector of AB is the negative reciprocal of $-\dfrac{3}{t}$, which is
$\dfrac{t}{3}.$
3. We know the perpendicular bisector of AB passes through the midpoint M$(t, 3)$. Hence, its equation in point-slope form is
$(y - 3) = \dfrac{t}{3}\,(x - t).$
Step 4: Find the Intersection of this Perpendicular Bisector with the y-axis
For a point on the y-axis, $x = 0.$
Substitute $x = 0$ into the line equation:
\[
(y - 3) = \dfrac{t}{3} (0 - t) = -\dfrac{t^2}{3}.
\]
Thus,
\[
y - 3 = -\dfrac{t^2}{3} \quad \Rightarrow \quad y = 3 - \dfrac{t^2}{3}.
\]
Therefore, the point $C$ on the y-axis is
\[
C\Bigl(0,\; 3 - \tfrac{t^2}{3}\Bigr).
\]
Step 5: Locate the Midpoint P of M and C
Let $P(h, k)$ be the midpoint of M $(t, 3)$ and C $\Bigl(0,\,3 - \tfrac{t^2}{3}\Bigr)$.
The midpoint coordinates are
\[
h = \dfrac{t + 0}{2} = \dfrac{t}{2},
\quad
k = \dfrac{3 + \left(3 - \tfrac{t^2}{3}\right)}{2}
= \dfrac{6 - \tfrac{t^2}{3}}{2}
= 3 - \dfrac{t^2}{6}.
\]
Step 6: Express the Coordinates of P in Terms of h and k
From $h = \dfrac{t}{2}$, we have $t = 2h.$
Substitute $t = 2h$ into the expression for $k$:
\[
k = 3 - \dfrac{(2h)^2}{6}
= 3 - \dfrac{4h^2}{6}
= 3 - \dfrac{2h^2}{3}.
\]
Thus,
\[
k = 3 - \dfrac{2}{3}h^2.
\]
Step 7: Derive the Locus Equation (Replace h by x and k by y)
Since $(h, k)$ is a general point on the locus, we write $x$ for $h$ and $y$ for $k$.
So,
\[
y = 3 - \dfrac{2}{3}x^2,
\]
or rearranging,
\[
\dfrac{2}{3}x^2 + y - 3 = 0 \quad \Longrightarrow \quad 2x^2 + 3y - 9 = 0.
\]
This is the required locus of the midpoint $P$.
Final Answer
The equation of the locus of $P$ is
\[
2x^2 + 3y - 9 = 0.
\]
