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How can I determine which is the directrix and the focus of a parabola and what is the distance between those points, only knowing that this parabola has its symmetry axis = OX and its passes through the points P1 and P2?

EDIT:

Guys, if possible, someone post an example using real numbers please, I think it will be more clear to me (both answers are great, but I'm still having problems to understand, sorry). I'm working with points P1(0,0) P2(6,6) and I need to found p (distance between directrix and focus). Can you guys explain using these numbers?

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  • $\begingroup$ Write the parabola as the graph of a polynomial of degree at most two. The three conditions on the parabola, that it pass through two points, and the condition on the symmetry axis, give three equations in the three unknown coefficients. Solve the resulting equations to find the coefficients. Then determine the directrix and the focus. $\endgroup$ – Dan Fox Aug 24 '15 at 13:44
  • $\begingroup$ Could you provide an answer exemplifying what you said for the points P1(0,0) and P2(6,6)? $\endgroup$ – chr0x Aug 24 '15 at 14:32
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Let's use the following form for the parabola:

$$y=a(x-h)^2+k$$

Distance from vertex $(h,k)$ to your directrix and focus is calculated using the following formula where $p$ is the distance.

$$a=\frac{1}{4p}$$

If $a$ is positive, the equation for your directrix will be as follows:

$$y=k-p$$

Also, the coordinates of the focus will be the following with $+a$:

$$(h,k+p)$$

If $a$ is negative, there are simply a few sign changes. Directrix equation with $-a$:

$$y=k+p$$

Focus with $-a$:

$$(h, k-p)$$

Hopefully this helped.

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  • $\begingroup$ Could you give me an example of how to find p using the points P1(0,0) and P2(6,6)? $\endgroup$ – chr0x Aug 24 '15 at 16:52
  • $\begingroup$ Let's assume $(0,0)$ is your vertex. This means $h=0$ and $k=0$, and your equation is the following: $$y=a(x-[0])^2 + [0]$$ Plugging in $6$ for $x$ and $6$ for $y$ (from point $(6,6)$), we get the following: $$6=a(6-0)^2+0$$ $$6=36a$$ $$a=\frac{1}{6}$$ Now plug $a$ into the following formula to get $p$: $$a=\frac{1}{4p}$$ $$p=1.5$$ You already know the following from earlier: $$h=0$$ $$k=0$$ $$a=\frac{1}{6}$$ Now use the formulae is listed earlier to solve for everything (in your case, $a$ is positive, so use those formulae. $\endgroup$ – Lanier Freeman Aug 24 '15 at 23:28
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The standard equation of the parabola with its axis (symmetric axis=OX) coincident with the x-axis & vertex at the point $(k, 0)$ on the x-axis is $$\color{blue}{y^2=4a(x-x_1)}$$

Where, $a$ & $k$ are arbitrary constants.

Now, satisfying the above equation of parabola by the coordinates of two given point $P_1(x_1, y_1)$ & $P_2(x_2, y_2)$. We get two linear equations in terms of $a$ & $k$ which are determined by solving them. Then we have

The equation of the directrix: $$x-k=-a\iff \color{red}{x=k-a}$$

The focus of the parabola: $$(x-k=a, y=0)\equiv\color{red}{(k+a, 0)}$$

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  • $\begingroup$ Hi Harish! I didn't get very well the formulas. You said that the equation: yˆ2=4a(x−x1) is "the standard equation of the parabola with vertex (x1, 0)". But, how can I assume that the vertex is on y=0 ? Another thing: x1 and x in that case are the x's relative to P1 and P2? So, if I have P1(0, 0) and P2(3,3) for example, the focus will be (3, 0)? Last but not least, could you provide me some references so I can complement my studies? I didn't found these formula in the books until now. $\endgroup$ – chr0x Aug 24 '15 at 14:04
  • $\begingroup$ Notice, the x-axis is the axis of the parabola hence its vertex will lie on the x-axis so we assume vertex at $(k, 0)$. For clarity, I have taken $k$ in place of $x_1$. You may check it in the edited answer. $\endgroup$ – Harish Chandra Rajpoot Aug 24 '15 at 15:38
  • $\begingroup$ Harish, I found this link: [1] here, and based on that, and using P1(0,0), P2(6,6) I found x=3 and k=-3. Could you tell me if thats right according to your answer? According to your answer I'll get a = 0, right? [1] algebra.com/algebra/homework/… $\endgroup$ – chr0x Aug 24 '15 at 17:36
  • $\begingroup$ Mr. Harish. Sadly I did not understood the meaning of variables that you used neither how can I apply the values I have, to solve the problem. Could you please add this information in your answer or add some reference that explains these informations? That way I can select your answer as the best answer and close this topic. Thanks. $\endgroup$ – chr0x Aug 24 '15 at 19:47

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