# Help with the proof of the Witch of Agnesi curve

$a=1$ (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!

• 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? – DonAntonio Mar 4 '14 at 19:47
• Hey Don! Yes because I tried few times but I didn't succeed..:-( – CS1 Mar 4 '14 at 19:49
• @DonAntonio, why you began with $x^2+4$? – CS1 Mar 4 '14 at 20:05
• Because I know where to go, @Yoav. You are asked to prove that relation, not to invent it. – DonAntonio Mar 4 '14 at 20:07
• Oh! I see!! Thank you @DonAntonio! it's the simplest prove here! Thank you so much!! – CS1 Mar 4 '14 at 20:17

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 $((4+x_P^2)y_B-8)y_P=0$.

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

Hint:

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

$x_A=x_P=x$,

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

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