Key Concepts and Formulas

• Midpoint Formula: The midpoint of a line segment joining points $(x_1, y_1)$ and $(x_2, y_2)$ is $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$.
• Slope of a Line: The slope of a line passing through points $(x_1, y_1)$ and $(x_2, y_2)$ is $m = \frac{y_2-y_1}{x_2-x_1}$.
• Perpendicular Lines: If two lines are perpendicular, the product of their slopes is -1 (i.e., $m_1 m_2 = -1$).
• Equation of a Line: The equation of a line with slope $m$ passing through point $(x_1, y_1)$ is $y - y_1 = m(x - x_1)$.

Step-by-Step Solution

Step 1: Find the coordinates of M, the midpoint of AB.

Given A(0, 6) and B(2t, 0), the midpoint M is: $M = \left(\frac{0+2t}{2}, \frac{6+0}{2}\right) = (t, 3)$

Step 2: Find the slope of AB.

The slope of the line AB is: $m_{AB} = \frac{0-6}{2t-0} = \frac{-6}{2t} = -\frac{3}{t}$

Step 3: Find the slope of the perpendicular bisector of AB.

The slope of the perpendicular bisector is the negative reciprocal of the slope of AB: $m_{\perp} = -\frac{1}{m_{AB}} = -\frac{1}{-\frac{3}{t}} = \frac{t}{3}$

Step 4: Find the equation of the perpendicular bisector of AB.

The perpendicular bisector passes through M(t, 3) and has a slope of $\frac{t}{3}$. Therefore, its equation is: $y - 3 = \frac{t}{3}(x - t)$ $3(y - 3) = t(x - t)$ $3y - 9 = tx - t^2$

Step 5: Find the coordinates of C, the point where the perpendicular bisector intersects the y-axis.

Since C lies on the y-axis, its x-coordinate is 0. Substitute x = 0 into the equation of the perpendicular bisector: $3y - 9 = t(0) - t^2$ $3y = 9 - t^2$ $y = \frac{9 - t^2}{3}$ So, the coordinates of C are $\left(0, \frac{9 - t^2}{3}\right)$.

Step 6: Find the coordinates of P, the midpoint of MC.

Given M(t, 3) and $C\left(0, \frac{9 - t^2}{3}\right)$, the midpoint P is: $P = \left(\frac{t+0}{2}, \frac{3 + \frac{9 - t^2}{3}}{2}\right) = \left(\frac{t}{2}, \frac{\frac{9 + 9 - t^2}{3}}{2}\right) = \left(\frac{t}{2}, \frac{18 - t^2}{6}\right)$

Step 7: Find the locus of P.

Let P = (x, y). Then we have: $x = \frac{t}{2} \implies t = 2x$ $y = \frac{18 - t^2}{6}$ Substitute t = 2x into the equation for y: $y = \frac{18 - (2x)^2}{6} = \frac{18 - 4x^2}{6} = 3 - \frac{2x^2}{3}$ $y = 3 - \frac{2}{3}x^2$ Multiply by 3: $3y = 9 - 2x^2$ $2x^2 = 9 - 3y$ $2x^2 + 3y - 9 = 0$ $2x^2 = 9-3y$

Step 8: Compare to the given options. The derived equation is $2x^2 + 3y - 9 = 0$. This does not match the 'Correct Answer' given. Let's re-examine our work.

Step 6 (Corrected): Find the coordinates of P, the midpoint of MC.

Given M(t, 3) and $C\left(0, \frac{9 - t^2}{3}\right)$, the midpoint P is: $P = \left(\frac{t+0}{2}, \frac{3 + \frac{9 - t^2}{3}}{2}\right) = \left(\frac{t}{2}, \frac{\frac{9 + 9 - t^2}{3}}{2}\right) = \left(\frac{t}{2}, \frac{18 - t^2}{6}\right)$

Step 7 (Corrected): Find the locus of P.

Let P = (x, y). Then we have: $x = \frac{t}{2} \implies t = 2x$ $y = \frac{18 - t^2}{6}$ Substitute t = 2x into the equation for y: $y = \frac{18 - (2x)^2}{6} = \frac{18 - 4x^2}{6} = 3 - \frac{2}{3}x^2$ $y - 3 = -\frac{2}{3}x^2$ Multiply by 3: $3y - 9 = -2x^2$ $2x^2 + 3y - 9 = 0 \implies 2x^2 = 9-3y$ This is still not the given answer. Let's re-examine the question and answer choices. The correct answer must be $3x^2 - 2y - 6 = 0$.

Since we are given that $3x^2 - 2y - 6 = 0$ is the correct answer, let's solve for y:

$2y = 3x^2 - 6$ $y = \frac{3}{2}x^2 - 3$

Since $x = \frac{t}{2}$, $t = 2x$. Substituting into $y = \frac{18 - t^2}{6}$:

$y = \frac{18 - (2x)^2}{6}$ $y = \frac{18 - 4x^2}{6} = 3 - \frac{2}{3}x^2$ $y = \frac{9 - 2x^2}{3}$

The error is in assuming the "Correct Answer" is incorrect. I will work backwards to find my mistake. Given that the correct answer is $3x^2 - 2y - 6 = 0$, let's express y in terms of x: $2y = 3x^2 - 6$ or $y = \frac{3}{2}x^2 - 3$.

We have $x = \frac{t}{2}$ and $y = \frac{18 - t^2}{6}$. Substitute $t = 2x$ into the expression for y:

$y = \frac{18 - (2x)^2}{6} = \frac{18 - 4x^2}{6} = 3 - \frac{2}{3}x^2$ $y = 3 - \frac{2}{3}x^2$ $3y = 9 - 2x^2$ $2x^2 = 9 - 3y$ $x^2 = \frac{9 - 3y}{2}$

The correct answer given is $3x^2 - 2y - 6 = 0$, which can be rewritten as $2y = 3x^2 - 6$. From $y = \frac{18 - t^2}{6}$, we get $6y = 18 - t^2$. Substituting $t = 2x$: $6y = 18 - 4x^2$. Dividing by 2, we have $3y = 9 - 2x^2$, or $2x^2 = 9 - 3y$.

Multiply the equation $3x^2 - 2y - 6 = 0$ by $\frac{2}{3}$ to get: $2x^2 - \frac{4}{3}y - 4 = 0$.

It appears there is an error in the problem statement or the provided correct answer. My calculations are consistent and do not yield the given answer.

Common Mistakes & Tips

• Double-check the midpoint formula and slope calculations.
• Remember that the slope of a perpendicular line is the negative reciprocal.
• Be careful when substituting to find the locus; make sure you express everything in terms of x and y.

Summary

We first found the midpoint M of AB. Then, we calculated the slope of AB and the slope of the perpendicular bisector. We found the equation of the perpendicular bisector and the coordinates of C. Next, we found the midpoint P of MC and, by eliminating the parameter t, derived the locus of P. After careful calculations, the locus of P is $2x^2 + 3y - 9 = 0$. However, this result does not match the options. Upon further review, it appears that none of the options are correct.

Final Answer

The derived equation $2x^2 + 3y - 9 = 0$ does not match any of the answer choices. There appears to be an error in the problem statement or the given correct answer. The closest option, if we made a sign error, is (A) $3x^2 - 2y - 6 = 0$. However, I cannot arrive at this answer through correct derivation.