Let P be the point (2, 0) and Q be a variable | vedclass

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(D) Let the coordinates of point $Q$ on the parabola $(y - 6)^2 = 2(x - 4)$ be $(4 + \frac{t^2}{2}, 6 + t)$.Given $P = (2, 0)$.Let the mid-point of $P

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$y^2 - x - 6y + 12 = 0$

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(D) Let the coordinates of point $Q$ on the parabola $(y - 6)^2 = 2(x - 4)$ be $(4 + \frac{t^2}{2}, 6 + t)$.Given $P = (2, 0)$.Let the mid-point of $PQ$ be $R(h, k)$.Then $h = \frac{2 + 4 + \frac{t^2}{2}}{2} = 3 + \frac{t^2}{4}$ and $k = \frac{0 + 6 + t}{2} = 3 + \frac{t}{2}$.From the second equation, $\frac{t}{2} = k - 3$, so $t = 2(k - 3)$.Substituting $t$ into the equation for $h$:$h = 3 + \frac{(2(k - 3))^2}{4} = 3 + \frac{4(k - 3)^2}{4} = 3 + (k - 3)^2$.$h - 3 = (k - 3)^2$.Replacing $(h, k)$ with $(x, y)$, we get $(y - 3)^2 = x - 3$.$y^2 - 6y + 9 = x - 3$.$y^2 - 6y - x + 12 = 0$.

Let $P$ be the point $(2, 0)$ and $Q$ be a variable point on the parabola $(y - 6)^2 = 2(x - 4)$. Then the locus of the mid-point of $PQ$ is:

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