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In an exercise I'm attempting to do, I am given the focus $(0, 3)$, the directrix ($x = 6$), and the eccentricity (e = $\frac{2}{3}$) and I'm being asked to obtain the conic equation. I don't know whether the conic is an ellipse, or a parabola or what. How can I approach this problem?

What I have been able to determine is, from the eccentricity, since e < 1, then the conic is actually an ellipse, right? My next step is to calculate a, b, c, from the eccentricity. Since it's given that $e=\frac{2}{3}$, and $e=\frac{c}{a}$, it seems simple enough to say that $c=2$, $a=3$ and then determine $b$ from Pitagoras' theorem. But I am unsure, as $c=4$, $a=6$ would also give the same eccentricity value. How would be the way to continue?

Also, given that the directrix is $x=6$, and the focus given is to the left side of the directrix, I'd guess that the focus I was given is the right focus on an horizontal ellipse. Since the eccentricity is the distance between centre and foci, I'd say that the centre is located at $(0-\frac{2}{3}, 3)$ and the other focus is at $(0-\frac{2}{3}-\frac{2}{3}, 3)$. But I'm not entirely sure I'm allowed to do that, is this the correct approach?

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2 Answers 2

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From definition of eccentricity,

\begin{align} \frac{2}{3} &= \frac{\sqrt{(x-0)^2+(y-3)^2}}{|x-6|} \\ \frac{4}{9} &= \frac{(x-0)^2+(y-3)^2}{(x-6)^2} \\ 4(x-6)^2 &= 9x^2+9(y-3)^2 \\ 4x^2-48x+144 &= 9x^2+9y^2-54y+81 \\ 0 &= 5x^2+9y^2+48x-54y-63 \end{align}

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  • $\begingroup$ Thanks, it turned out to be way simpler than I thought... $\endgroup$ Commented May 29, 2018 at 19:46
  • $\begingroup$ a follow-up question, I suppose this could be used regardless of the conic type (i.e. a parabola)? $\endgroup$ Commented May 29, 2018 at 19:57
  • $\begingroup$ Answering my own comment, it seems so. I tried for $f(-2,3)$, $x=5$ and e=1, and got $y^2+14x-6y-12=0$, which is a horizontal parabola. $\endgroup$ Commented May 29, 2018 at 20:16
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It appears that you’re way overthinking this. One definition of a conic is the locus of points for which the ratio of their distances to a fixed point (the focus) and fixed line (the directrix) is constant (the eccentricity). Using this definition and standard formulas for the distance between two points and the distance from a point to a line you can write an equation down directly.

Let the coordinates of the focus be $F=(x_f,y_f)$, the equation of the directrix be $ax+by+c=0$ and the eccentricity be $e$. According to the above definition, then, points on the curve satisfy the equation $$\sqrt{(x-x_f)^2+(y-y_f)^2} = e{|ax+by+c|\over\sqrt{a^2+b^2}}.$$ This is a bit inconvenient to work with, but you can square both sides to get $$(x-x_f)^2+(y-y_f)^2 = e^2{(ax+by+c)^2\over a^2+b^2}.$$ Rearrange and simplify as you see fit.

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  • $\begingroup$ I see thanks, it seems that, as you say, I was overthinking it too much. thanks! $\endgroup$ Commented May 29, 2018 at 19:45

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