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Given the vertex of a parabola $A(-2,-1)$ and the equation of its directrix $x+2y-1=0$ find the equation of this parabola. I send my procedure, I just need to write the equation of the parabola, I tried to equate $PF = PA$ but I remove the $x$ and $y$ square. Thank you very much.

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  • $\begingroup$ By ''guideline'' you means the directrix? $\endgroup$ – Emilio Novati Apr 13 '17 at 19:54
  • $\begingroup$ Yes @EmilioNovati $\endgroup$ – p472558 Apr 13 '17 at 19:57
  • $\begingroup$ See math.stackexchange.com/a/1429421/265466 for an answer to a similar question (not coincidentally, written by @EmilioNovati). $\endgroup$ – amd Apr 13 '17 at 20:17
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Hint:

1) find the line $s$ orthogonal to the directrix $d$ an passing thorough $A$ ( this is the symmetry axis of the parabola)

2)find the point $D$ that is the intersection of $r$ and $d$.

3) fine the focus of the parabola that is the point $F$ on the axis such that $\overline{FA}=\overline{AD}$

4) now you can write the equation of the parabola as the locus of points equdistant from $F$ and $d$ ( see here)

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What you have done so far is fine, and you have the coordinates of the focus as $(-3,-3).$

Therefore using the focus-directrix property of the ellipse, and the formula for the distance from a point to a line, the equation of the parabola is $$\frac{(x+2y-1)^2}{5}=(x+3)^2+(y+3)^2$$

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What you’ve done so far is good. You just need to complete the calculation by using the focus-directrix property of a parabola and using the squared distances, as David Quinn describes in his answer. I’d like to show you a different way to find the focus using vectors that might be a little less work than what you did.

From the equation of the directrix we can read that the vector $\vec n=\langle1,2\rangle$ is normal (perpendicular) to it. The formula for the signed distance of a point from a line gives us $$-{-2+2(-1)-1\over\sqrt{1^2+2^2}}=\sqrt5$$ for the distance from $A$ to the directrix. That is, if we go $\sqrt5$ units in the direction of $\vec n$ from $A$ to the point $A+\sqrt5{\vec n\over\|\vec n\|}$ we reach the directrix. Since the vertex is halfway between the focus and directrix, this means that the focus $F$ is the same distance in the other direction, therefore $$F=A-\sqrt5{\vec n\over\|\vec n\|}=(-2,-1)-\sqrt5{\langle1,2\rangle\over\sqrt5}=(-3,-3).$$ (The above divisions by the length of $\vec n$ are in order to get a unit vector in the same direction.)

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