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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?

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  • 1
    $\begingroup$ 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. $\endgroup$ – amd Aug 25 '19 at 0:52
  • $\begingroup$ 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. $\endgroup$ – amd Aug 25 '19 at 0:54
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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?

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  • $\begingroup$ 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 $C$. $\endgroup$ – Aleksandra Asanin Aug 25 '19 at 14:18
  • $\begingroup$ Even if you are right, then the final solution would be $4(x_P+4)$ which again mismatches with the book solution. :) $\endgroup$ – Aleksandra Asanin Aug 25 '19 at 18:35
  • $\begingroup$ Perfect... Thanks! $\endgroup$ – Aleksandra Asanin Aug 25 '19 at 19:17
  • $\begingroup$ You are welcome $\endgroup$ – Luca Goldoni Ph.D. Aug 25 '19 at 19:17

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