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Hello I have tried to post a few questions to try to understand better some things in conics. I am still really confused and am looking for help to understand. I am especially confused about graphing conics and the relation between the directrix and eccentricity. Maybe the best way for me to explain is give the examples of some of the questions asked and show what I have tried and what I am confused about.

One of the questions says given a focus of (5,0) and a directrix of x=0 draw a conic with eccentricity of , then it gives a few options

1) 0.2

For this I just used that $\frac{PF}{PD}=\frac{1}{5}$ so $5PF=PD$ My thoughts are this is an ellipse.

But I dont know where to go from here, if I were to say x is the distance from the directrix to the left vertex that I could say $PD=x$ then $PF=x/5$ for example but I dont know.

2) 0.8

also an ellipse and similar to above but $\frac{PF}{PD}=\frac{4}{5}$ instead. But I dont know how I can find points of these?

3) 1

I know this is supposed to be a parabola. With bassicaly just PF=PD , is that all I can say? How could I actually find a point? How can I find the origin?

4) 5

$$\frac{PF}{PD}=5$$

I know this is supposed to be a Hyperbola with $PF=5PD$ but after that im lost.

I really hope someone can take the time to help me understand. Thanks a lot for anyone who can help

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Hint.

For $X=(x,y)$

the equation of the conic is $ \overline {XF} = e \overline {Xd} \qquad (1)$

where:

$e$ is the eccentricity, $\overline {XF}$ is the distance from the focus, that in your case is $\sqrt{(x-5)^2+y^2}$ and $\overline {Xd}$ is the distance from the directix, that in your case is $|x|$. (see).

Write the equation $(1)$ for the differnt eccentricities , square (and reorder), and you have the equations of the conics.

As examples:

for $e=\frac{1}{5}$ ( your case 1 ), you find the equation:

$$ (x-5)^2+y^2=\dfrac{1}{25}x^2 $$ Can you show that this is an ellipse?

For $e=1$ ( case 3) the equation is: $$ (x-5)^2+y^2=x^2 $$ That is a parabola.

... and so one on the other cases...

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  • $\begingroup$ You are reffering to example (1)? And that equation holds for all conic sections? $\endgroup$ – Quality Oct 11 '15 at 20:21
  • $\begingroup$ I'm referring to $ \overline {XF} = e \overline {Xd} \qquad (1)$ and, yes, this equation holds for all conics. $\endgroup$ – Emilio Novati Oct 11 '15 at 20:27
  • $\begingroup$ Thank you, I mean when you say in your case, but I put a few cases $\endgroup$ – Quality Oct 11 '15 at 20:28
  • $\begingroup$ This works in all your cases, simply use the value of the eccentricity $e$ that you want. $\endgroup$ – Emilio Novati Oct 11 '15 at 20:31
  • $\begingroup$ @Quality: I've added to my answer. I hope it's usefull ;) $\endgroup$ – Emilio Novati Oct 11 '15 at 20:53

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