# 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?

• Well, to begin with, your simplified expression for the distance to the line has the wrong sign: try $x_P=0$ in it. That aside, halving this distance doesn’t give you the $x$-coordinate of the midpoint. It gives you the distance of the midpoint from the line. – amd Aug 25 '19 at 0:52
• You don’t need to compute these distances, anyway. For any two points $P$ and $Q$, the coordinates of their midpoint is $(P+Q)/2$, and the foot of the perpendicular from $(x_P,y_P)$ to the line $x+3=0$ is just $(-3,y_P)$ since, as you’ve noted, this perpendicular is horizontal. – amd Aug 25 '19 at 0:54

$$\frac{{x_P + x_Q }} {2} = x_M$$ 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?
• But the book says that the sign before $\sqrt{A^2+B^2}$ is opposite of $C$ if $C \neq 0$, which is, in this case, a negative sign due to positive $C$. – Aleksandra Asanin Aug 25 '19 at 14:18
• Even if you are right, then the final solution would be $4(x_P+4)$ which again mismatches with the book solution. :) – Aleksandra Asanin Aug 25 '19 at 18:35