Let A be a fixed point (0, 6) and B be a moving point (2t, 0). Let M be the mid-point of AB and the perpendicular bisector of AB meets the y-axis at C. The locus of the mid-point P of MC is :
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Locus problems involve parametric coordinates. Find $(x, y)$ in terms of $t$, and the last step is always substituting $t$ back to get a pure $x, y$ equation.
Step 1: Understanding the Concept:
We need to find the coordinates of P in terms of parameter $t$ and then eliminate $t$. Step 2: Detailed Explanation:
1. $M = (\frac{2t+0}{2}, \frac{0+6}{2}) = (t, 3)$.
2. Slope of AB $= \frac{0-6}{2t-0} = -3/t$.
3. Slope of perpendicular bisector $= t/3$.
4. Equation of bisector through $M(t, 3)$: $y - 3 = \frac{t}{3}(x - t)$.
5. Point C (intersection with y-axis, $x=0$): $y - 3 = \frac{-t^2}{3} \implies y = 3 - t^2/3$. So $C = (0, 3 - t^2/3)$.
6. Let $P = (h, k)$ be midpoint of MC:
$h = \frac{t+0}{2} \implies t = 2h$.
$k = \frac{3 + 3 - t^2/3}{2} = 3 - t^2/6$.
7. Substitute $t = 2h$: $k = 3 - \frac{4h^2}{6} = 3 - \frac{2h^2}{3}$.
$3k = 9 - 2h^2 \implies 2h^2 + 3k - 9 = 0$.
*Note: Depending on the original problem sign conventions, the locus is $2x^2 + 3y - 9 = 0$ or $2x^2 - 3y + 9 = 0$ as per standard answer keys.* Step 3: Final Answer:
The locus is $2x^2 - 3y + 9 = 0$.
