Let A be a fixed point (0,6) and B be a moving point (2t,0).

Question

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:

Accepted Answer

$M=(t,3)$. Perpendicular bisector of $AB$ (slope of $AB=\dfrac{-6}{2t}=-\dfrac3t$, so perp. slope $=\dfrac t3$) through $M$:
$$y-3=\frac t3(x-t)$$

At $x=0$: $y = 3-\dfrac{t^2}3$, so $C=\left(0,3-\dfrac{t^2}3\right)$.

Midpoint $P(h,k)$ of $M(t,3)$ and $C\left(0,3-\dfrac{t^2}3\right)$:
$$h=\frac t2, \qquad k = 3-\frac{t^2}6$$

From $h=\dfrac t2$: $t=2h$. Substituting: $k=3-\dfrac{4h^2}6 \implies 6k=18-4h^2\cdot\dfrac{6}{6}$... simplify: $k=3-\dfrac{2h^2}3 \implies 3k=9-2h^2 \implies 2h^2+3k-9=0$

Locus: $2x^2+3y-9=0$.

Solutions · 1