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I want to solve this problem. I know a focus of a parabola: $F=(7,-3)$ and 2 points which parabola lies on $A=(5,7/3)$ and $B=(5,-25/3)$. I also know y-coordinate of vertex of a parabola: $V=(?,-3)$. I have to find vertex equation of a parabola. Thanks for your help.

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    $\begingroup$ Your parabola should would be of the form $x=A(y+3)^2+7 $and then you solve by substituting your two points A and B. (Actually one would suffice.) Then you substitute your value for $y$ in $V$. $\endgroup$ – daruma Jul 28 '16 at 11:36
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    $\begingroup$ The parabola $(y-n)^2=4p(x-m)$ has the vertex $(m,n)$, the axis of symmetry $y=n$ and the focus $(p+m,n)$. $\endgroup$ – mathlove Jul 28 '16 at 11:49
  • $\begingroup$ How can I get a value of $p$? $\endgroup$ – T. Böhm Jul 28 '16 at 11:51
  • $\begingroup$ @T.Böhm: We have $n=-3$ and $p+m=7$. Now substitute $x=5,y=7/3$. $\endgroup$ – mathlove Jul 28 '16 at 11:54
  • $\begingroup$ @daruma: I don't think so. $(7,-3)$ is the focus, not the vertex. $\endgroup$ – mathlove Jul 28 '16 at 11:57
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Two points on a parabola and its focus just fix a pair of parabolas, and we cannot simply assume that their axis will be parallel/perpendicular to the $y$-axis. Let $A,B$ be our points, $F$ our focus. Let $A',B'$ be the projections of $A,B$ on the directrix: clearly, $A'$ belongs to the circle $\Gamma_A$ centered at $A$ through $F$, $B'$ belongs to the circle $\Gamma_B$ centered at $B$ through $F$. The directrix is a common tangent to $\Gamma_A$, $\Gamma_B$, so it goes through the exterior homothetic centre $H$ of $\Gamma_A,\Gamma_B$. $H$ obviously lies on $AB$, and its position on the $AB$ line just depends on the radii of $\Gamma_A$, $\Gamma_B$, i.e. $AF$ and $BF$.

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Summarizing:

  1. Find $AF,BF$ through the Pythagorean theorem;
  2. Use the previous informations to locate $H\in AB$;
  3. One of the tangents to $\Gamma_A$ through $A$ is the directrix of the parabola;
  4. Given the focus and the directrix, there is nothing else to locate: the vertex is just the midpoint of the segment joining $F$ with its projection on the directrix.
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  • $\begingroup$ Sorry to bother you, I want to ask you something. My understanding was that having $F(7,-3), V(?,-3)$ leads that the axis of symmetry is $y=-3$ which is perpendicular to $y$-axis. It seems that this is wrong according to your answer. Can you please tell me why? $\endgroup$ – mathlove Jul 29 '16 at 5:21
  • $\begingroup$ @mathlove: if some point lies on the line y=3 it doesn't mean that the whole axis has be such a line... $\endgroup$ – Jack D'Aurizio Jul 29 '16 at 8:41
  • $\begingroup$ Well, yes, but here $F$ is the focus, $V$ is the vertex. So, my understanding is that the line passing through $F$ and $V$ is the axis of symmetry. Is this wrong? $\endgroup$ – mathlove Jul 29 '16 at 8:44
  • $\begingroup$ No, but we are stating and using different things. In particular my solution is more general because it does not need any info on V to work, while you are deeply relying on the fact that F and V have the same y coordinate. $\endgroup$ – Jack D'Aurizio Jul 29 '16 at 8:45
  • $\begingroup$ Yes, I am relying on the fact because the OP says that "I also know y-coordinate of vertex of a parabola: $V=(?,-3)$." $\endgroup$ – mathlove Jul 29 '16 at 8:46

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