Q:

Geometry question mismatch with the book solution

Question: A point $M$ moves on the curve $y^2 = 8x + 4$. A line $L$ passes through $M$ and is perpendicular to the line $x+3=0$, the foot of the perpendicular is $Q$. If $M$ is the midpoint of $PQ$, find the equation of the locus of $P$.
What I did:
Distance between the point $P$ and the line $L$ is: 
$\dfrac{Ax_p + By_p+C}{\sqrt{A^2+B^2}}$, where $A=1, B=0, C=3$ 
according to the equation of $L$, and the sign of $\sqrt{A^2+B^2}$ is the opposite of $C$, so it is negative.
Finally, the distance between $L$ and $P$ is $-(x_P+3)$.
Since the point $M$ is the midpoint of $PQ$, which is a horizontal line (obviously because it is perpendicular to the vertical line $x+3=0$), then the $x$-coordinate of $M$ would be $x_M = \dfrac{-(x_P+3)}{2}$.
If now $x$-coordinate is replaced in the curve equation, we get that:
${y_M}^2 = - 8 \left(  \dfrac{x_P+3}{2} \right) + 4 = -4x_P - 12 + 4  = 4(-x_P-2) $.
But the solution in the book is $4(x_P-2)$. 
What am I doing wrong here?

A:

Try this
$$
\frac{{x_P  + x_Q }}
{2} = x_M 
$$
thus
$$
\frac{{x_P  - 3}}
{2} = x_M 
$$
Now
$$
y^2 _M  = 8x_M  + 4
$$
and since $$y_M=y_P$$ you have
$$
y^2 _P  = 8\left( {\frac{{x_P  - 3}}
{2}} \right) + 4
$$
Therefore
$$
y^2 _P  = 4x_P  - 8 = 4\left( {x_P  - 2} \right)
$$
Thus
$$
y^2 _P  = 4\left( {x_P  - 2} \right)
$$
As as in your book. It's clear now?

