Parametrics Domain and Range Question

[Jefferson]

Hi everyone, please help me with the following parametric questions. Thank you.

P (2ap , ap²) and Q (2aq , aq²), where a > 0 and pq = 1, are two points on the parabola x² = 4ay.

M is the midpoint of PQ.
M: (  a(p+q) , 0.5a(p² + q²)    )

As P and Q move on the parabola, the locus of M is:
x²  = 2a(y + a)

(those are absolute signs, | | ) Given that:
| p + 1/p | ≥ 2 for p ≠ 0,

Find the domain and range of the locus of M.

[RuiAce]

You can retrieve the domain the following way.
\[ \text{Of course, since }pq = 1,\text{ we know that }\boxed{q=\frac1p}.\\ \text{Therefore the equation for the }x\text{-coordinate of }M\text{ is}\\ \boxed{x = a\left(p+\frac1p\right)}. \]
\[ \text{So immediately we see that }\boxed{\frac{x}{a} = p+\frac{1}{p}}.\\ \text{Since we're given that }\left| p +\frac{1}{p}\right| \geq 2, \\ \text{we instantly know that}\\ \left| \frac{x}{a}\right| \geq 2. \]
The solution to this absolute value inequality is just \(x\leq -2a\) or \(x\geq 2a\). You should be able to retrieve the range from here.

(Which can be done so in a variety of ways. One brute force way is just to make \(y\) the subject of the actual equation of the locus, and then sub \( x^2 \geq 4a^2\) in.)