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$a=1$ (The radius is 1). How do I prove that if we talking about $P=(x,y)$, then: $$y=\frac{8}{x^2+4}$$ I'd like to get any help!

Thank you! enter image description here

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    $\begingroup$ You already asked about this a few hours ago and you were shown that the parametrics of the curve are $$x=2a\cot t\;,\;\;y=2a\sin^2t\implies\;\text{with $\;a=1\;$ we have:}$$ $$x^2+4=4\cot^2t+4=4\left(1+\frac{\cos^2t}{\sin^2t}\right)=\frac4{\sin^2t} \implies$$ $$\implies\frac8{x^2+4}=2\sin^2t=y$$ Did you really need to ask this question? $\endgroup$ – DonAntonio Mar 4 '14 at 19:47
  • $\begingroup$ Hey Don! Yes because I tried few times but I didn't succeed..:-( $\endgroup$ – CS1 Mar 4 '14 at 19:49
  • $\begingroup$ @DonAntonio, why you began with $x^2+4$? $\endgroup$ – CS1 Mar 4 '14 at 20:05
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    $\begingroup$ Because I know where to go, @Yoav. You are asked to prove that relation, not to invent it. $\endgroup$ – DonAntonio Mar 4 '14 at 20:07
  • $\begingroup$ Oh! I see!! Thank you @DonAntonio! it's the simplest prove here! Thank you so much!! $\endgroup$ – CS1 Mar 4 '14 at 20:17
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The line through $O$ and $A$ has equation $x=\frac{x_A}{2}y$, the circle has equation $x^2+(y-1)^2=1$. Substituting $x$ from the first into the second you find the coordinates $(x_B,y_B)$ of $B$. You get $(\frac{x_A}{2}y_B)^2+(y_B-1)^2=1$ whence $((4+x_A^2)y_B-8)y_B=0$. Noting that $y_P=y_B$ and $x_P=x_A$ you finally get $((4+x_P^2)y_B-8)y_P=0$.

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  • $\begingroup$ What do you mean when you write: "Substituting $x$ from the first into the second"? $\endgroup$ – CS1 Mar 4 '14 at 18:30
  • $\begingroup$ @Yoav Fridman Edited. Is it clear now? $\endgroup$ – alex Mar 4 '14 at 18:36
  • $\begingroup$ Alex, but how it's describe the coordinates of $P$? i.e. x,y should be the coordinates of $P$. $\endgroup$ – CS1 Mar 4 '14 at 19:38
  • $\begingroup$ @Yoav Fridman $y_B=y_P$, $x_A=x_P$. $\endgroup$ – alex Mar 4 '14 at 19:49
  • $\begingroup$ Oh...So $y$ is $y_B$? I'm right? $\endgroup$ – CS1 Mar 4 '14 at 19:53
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Hint:

$x_B^2=1-(1-y_B)^2$,

$x_A=x_P=x$,

$y_B=y_P=y$ and $$\frac{x_A}{y_A}=\frac{x_B}{y_B}$$

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  • $\begingroup$ And how I continue from there? Thank you! $\endgroup$ – CS1 Mar 4 '14 at 18:48
  • $\begingroup$ @YoavFridman, After substitution in fractions and squaring we have $$\frac{x^2}{4}=\frac{1-(1-y)^2}{y^2}=-1+\frac{2}{y}$$ and it gives us $$y=\frac{8}{x^2+4}$$ $\endgroup$ – Woria Mar 4 '14 at 19:42
  • $\begingroup$ How do you get the first line: $\frac{x^2}{4}=\frac{1-(1-y)^2}{y^2}=-1+\frac{2}{y}$? $\endgroup$ – CS1 Mar 4 '14 at 20:03
  • $\begingroup$ @YoavFridman, $x_A=x_P=x$, $y_A=2$, $x_B=(1-(1-y)^2)^{1/2}$, and $y_B=y_P=y$. $\endgroup$ – Woria Mar 4 '14 at 20:29

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