Locus of the midpoint of $(2,6)$ and a point on $y^2=2x$

Question

I need help with the following:

$B$ is any point on the curve $y^2=2x$ and $A$ is a fixed point with coordinates $(2,6)$. Find the equation of the curve traced out by the midpoint of $AB$.

The main problem I have is trying to figure how to derive the equation… and how to use coordinates to derive an equation.

Answer 1

Let $M(x,y)$ is a point in our curve and $B(x_0,y_0)$.

Hence. $x=\frac{x_0+2}{2}$ and $y=\frac{y_0+6}{2}$.

Thus, $(2y-6)^2=2(2x-2)$

Answer 2

Write the curve $y^2=2x$ in parametric form: $$(x,y)=\left(\frac{t^2}2,t\right)\quad t\in\Bbb R$$ Each value of $t$ is associated with a possible point for $B$. The midpoint of $AB$ can then be written in terms of $t$ as $$\left(\frac{t^2/2+2}2,\frac{t+6}2\right)=\left(\frac{t^2}4+1,\frac t2+3\right)$$ Substituting $y=\frac t2+3$ yields $$(x,y)=((y-3)^2+1,y)$$ or, in implicit terms, $x=(y-3)^2+1$. This is the equation of the midpoint of $AB$.