Circle in parabola

  • Thread starter Weave
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  • #1
Weave
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Homework Statement


A circle with a radius of 1inscribed in the parabola y=x^2, find the center of the circle. The figure shows the circle on the y-axis.

Homework Equations


[tex]y=x^2[/tex]
[tex] r^2=(x-h)^2+(y-k)^2[/tex]

The Attempt at a Solution


h=0. [tex]\frac{x^2+(y-k)^2}{r^2}=x^2[/tex]
R^2=1 and the x^2 can be subtracted out leaving:[tex](y-k)^2=0[/tex] take the derivitive of each side, find y, substitute that back in and find k but I end up with other varibles to solve.
 
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Answers and Replies

  • #2
d_leet
1,074
1

Homework Statement


A circle with a radius of 1inscribed in the parabola y=x^2, find the center of the circle. The figure shows the circle on the y-axis.

The Attempt at a Solution


h=0. [tex]\frac{x^2+(y-k)^2}{r^2}=x^2[/tex]

How did you get this equation?
 
  • #3
Weave
143
0
setting one equation equal to another
 
  • #4
d_leet
1,074
1
setting one equation equal to another

And how did you do that? Can you show some work because I cannot see what you set equal to what.
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
43,010
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The equation of the circle and the equation of the parabola are describing different curves. They do NOT always have the same (x,y) values!

Unfortunately, we can't see the figure you talk about. In what sense is the circle "inscribed" in the parabola? Is it tangent to the parabola at two points?

From symmetry, it should be obvious that the circle has center on the y-axis. That is, that the equation of the circle is x2+ (y- h)2= r2.

The derivative of y= x2 is y'= 2x and the derivative of y in x2+ (y-h)2= r2 is given by 2x+ 2(y- h)2y'= 0.

You have 4 equations: y= x2, x2+ (y-h)2= r2, y'= 2x, and 2x+ 2(y-h)2= 0 to solve for four unknown values: the (x,y) values of points of intersection, h, and r.
 

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