so right now I am taking Algebra 2, and we are learning about Conic Sections. We were of course given the functions to form the shapes of each section (circle, ellipse, parabola, hyperbola). I was just wondering why do these functions form these shapes when graphed? My teacher just gave us the functions, but didn't explain this. Could someone shed some light and/or tell me if this is a useless question? (Let me know if you need me to clarify anything, I'm not sure if I explained my question correctly.)
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I believe you either mean $$y^2=4cx$$ or $$x^2=4cy$$? – Karn Watcharasupat May 03 '18 at 19:24
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Have you learnt the definition of parabola? – Karn Watcharasupat May 03 '18 at 19:25
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I just learned the definition of a parabola today. Also, I don't know how to enter the equations like that on the forum. The equations you showed aren't what I was trying to enter. – CaptainAmerica16 May 03 '18 at 19:25
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What is the definition you learnt? (There are a few versions...which are ultimately equivalent) – Karn Watcharasupat May 03 '18 at 19:26
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MathJax tutorial: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Karn Watcharasupat May 03 '18 at 19:27
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A parabola is the set of all points in a plane that are the same distance from a fixed line (directrix) and a point not on the line (focus). – CaptainAmerica16 May 03 '18 at 19:27
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$$ y = \frac {(1)}{4c} x^2$$ – CaptainAmerica16 May 03 '18 at 19:33
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Ok, this is what I was trying to show. Should I edit the title as well? – CaptainAmerica16 May 03 '18 at 19:33
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I see. See my answer for the explanation. and yep you should. – Karn Watcharasupat May 03 '18 at 19:34
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Suppose we have the focus on the $x$-axis as $F(c,0)$ and the directrix $x=-c$.
Then from the definition, a point $P(x,y)$ on the parabola satisfies $$PF=Pd_\perp$$ thus $$\sqrt{(x-c)^2+(y-0)^2}=|x-(-c)|$$ Simplifying, $$y^2=(x+c)^2-(x-c)^2=4cx$$
Suppose we have the focus on the $y$-axis as $F(0,c)$ and the directrix $y=-c$.
Then from the definition, a point $P(x,y)$ on the parabola satisfies $$PF=Pd_\perp$$ thus $$\sqrt{(y-c)^2+(x-0)^2}=|y-(-c)|$$ Simplifying, $$x^2=(y+c)^2-(y-c)^2=4cy$$ which gives $$y=\frac{1}{4c}x^2$$
Karn Watcharasupat
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Thanks for putting time into this explanation. As I was looking over this and the definition I received in class, this seems like a bit of a silly question :) – CaptainAmerica16 May 03 '18 at 19:49
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1@CaptainAmerica16 That's alright haha I was quite confused w conics when I first started learning it also. It's very normal to not be able to link the algebraic and geometric sides of conics together at the start. – Karn Watcharasupat May 04 '18 at 12:01