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March 18th, 2014, 08:05 AM   #1
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Circle tangent to parabola

Hi,

I am stuck with some homework and I hope someone can help me. The problem is depicted in the attachment. A circle moves along a parabola. The purpose is to find the orbit of the center of the circle. The contact point of a circle and a parabola is $\({x_p},{y_p})\$ and the center of the circle is $\({x_m},{y_m})\$. r and F are respectively the radius of the circle and the focus of the parabola. I have to show that:
$\{x_m}= \frac{{{x_p}}}{{4F}} + \frac{{2Fr}}{{\sqrt {x_p^2 + 4{F^2}} }}\$
and that Xp is a solution of:
$\x= {x_m} + \frac{{rx}}{{\sqrt {{x^2} + 4{F^2}} }}\$
I am stuck and would really appreciate if someone can help me. Thanks in advance.
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March 18th, 2014, 08:32 AM   #2
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Re: Circle tangent to parabola

Quote:
 Originally Posted by Gluon I have to show that: $\{x_m}= \frac{{{x_p}}}{{4F}} + \frac{{2Fr}}{{\sqrt {x_p^2 + 4{F^2}} }}\$
Are you sure it's not

$y_m= \frac{{{x_p^2}}}{{4F}} + \frac{{2Fr}}{{\sqrt {x_p^2 + 4{F^2}} }}$ ?

Quote:
 Originally Posted by Gluon and that Xp is a solution of: $\x= {x_m} + \frac{{rx}}{{\sqrt {{x^2} + 4{F^2}} }}\$
Are you sure it's not

$x_p= x_m + \frac{{rx_p}}{{\sqrt {{x_p^2} + 4{F^2}} }}\$ ?

Do you know how to find a tangent vector at $(x_p,y_p)$ an then how to turn it 90 degrees counterclockwise?

March 18th, 2014, 09:16 AM   #3
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Re: Circle tangent to parabola

Quote:
Originally Posted by Evgeny.Makarov
Quote:
 Originally Posted by Gluon I have to show that: $\{x_m}= \frac{{{x_p}}}{{4F}} + \frac{{2Fr}}{{\sqrt {x_p^2 + 4{F^2}} }}\$
Are you sure it's not

$y_m= \frac{{{x_p^2}}}{{4F}} + \frac{{2Fr}}{{\sqrt {x_p^2 + 4{F^2}} }}$ ?

Quote:
 Originally Posted by Gluon and that Xp is a solution of: $\x= {x_m} + \frac{{rx}}{{\sqrt {{x^2} + 4{F^2}} }}\$
Are you sure it's not

$x_p= x_m + \frac{{rx_p}}{{\sqrt {{x_p^2} + 4{F^2}} }}\$ ?

Do you know how to find a tangent vector at $(x_p,y_p)$ an then how to turn it 90 degrees counterclockwise?
Yes, indeed you're right but I can't find the edit button to change my post. I know how to find a tangent line to a circle with implicit differentiation, but I don't know how to proceed. Any suggestions?

 March 18th, 2014, 09:30 AM #4 Senior Member   Joined: Dec 2013 From: Russia Posts: 327 Thanks: 108 Re: Circle tangent to parabola A tangent vector to the graph y = f(x) at point $x_0$ has slope $f'(x_0)$. Thus, as a tangent vector itself you can take $(1, f'(x_0))$ or any vector proportional to it. Next, if (a, b) is a vector, then the result of turning it by 90 degrees is (-b, a). Indeed, their scalar product is a(-b) + ba = 0, which means they are perpendicular. Finally, if $(x_m,y_m)$ is obtained by shifting $(x_p,y_p)$ in the direction of some vector (a, b) by distance r, then $x_m=x_p+a/\sqrt{a^2+b^2}$, and similarly for y. Indeed, $(a, b)/sqrt{a^2+b^2}$ is the unit vector pointing from $(x_p,y_p)$ to $(x_m,y_m)$. So, take $f(x)=x^2/(4F)$ and find some tangent vector at $(x_p,y_p)$ (preferably, its coordinates should not have fractions for easier calculations). Then find a perpendicular vector. Find a unit perpendicular vector. Multiply it by r and use the result to shift $(x_p,y_p)$ to $(x_m,y_m)$.
 March 18th, 2014, 10:28 AM #5 Newbie   Joined: Mar 2014 Posts: 4 Thanks: 1 Re: Circle tangent to parabola Wow great, thanks! 1. Tangent vector at $({x_p},{y_p})$ : $(1,\frac{{{x_p}}}{{2F}})= (2F,{x_p})$. 2. Perpendicular unit vector: $(\frac{{ - {x_p}}}{{\sqrt {x_p^2 + 4{F^2}} }},\frac{{2F}}{{\sqrt {x_p^2 + 4{F^2}} }})$ 3. Thus: $\begin{array}{l} {y_m} = {y_p} + \frac{{2F}}{{\sqrt {x_p^2 + 4{F^2}} }} \cdot r\\ {x_p} = {x_m} - \frac{{ - x}}{{\sqrt {x_p^2 + 4{F^2}} }} \cdot r = {x_m} + \frac{x}{{\sqrt {x^2 + 4{F^2}} }} \cdot r \end{array}$ Should be good, right?
March 18th, 2014, 10:38 AM   #6
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Re: Circle tangent to parabola

Quote:
 Originally Posted by Gluon 1. Tangent vector at $({x_p},{y_p})$ : $(1,\frac{{{x_p}}}{{2F}})= (2F,{x_p})$.
$(1,\frac{{{x_p}}}{{2F}})$ and $(2F,{x_p})$ are parallel vectors, but they are not equal.
Quote:
 Originally Posted by Gluon 3. Thus: $\begin{array}{l} {y_m} = {y_p} + \frac{{2F}}{{\sqrt {x_p^2 + 4{F^2}} }} \cdot r\\ {x_p} = {x_m} - \frac{{ - x}}{{\sqrt {x_p^2 + 4{F^2}} }} \cdot r = {x_m} + \frac{x}{{\sqrt {x^2 + 4{F^2}} }} \cdot r \end{array}$
In the second equation, x should be replaced by $x_p$. Otherwise, this is correct.

 March 18th, 2014, 11:21 AM #7 Newbie   Joined: Mar 2014 Posts: 4 Thanks: 1 Re: Circle tangent to parabola Ok I get it. Thanks for the effort and help

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