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A circle rests in the interior of the parabola with an equation y=x^2 so that it is tangent to the parabola at two points. How much higher is the center of the circle than the points of tangency?

 

Thank you very much!

:P

 Dec 29, 2018
 #1
avatar+18965 
0

Is there any more info given for this question.......like the size or equation for the circle?  I think there are MANY circles that would fit the bill.....do you need an answer in terms of the circle size/equation?

 Dec 29, 2018
 #5
avatar+800 
-1

Sorry for answering late...

 

CPhill already got the answer. If you could look at his, you may understand :)

CoolStuffYT  Jan 1, 2019
 #2
avatar+103131 
+2

Because of symmetry, the center of the circle will lie  at  (0, k)

 

Let the equation of the  circle be

 

x^2 + (y - k)^2 = r^2

x^2 + y^2 - 2ky + k^2 = r^2

Using implicit differentiation, the slope of any tangent line to the circle is given by

 

2x + 2yy' - 2ky'  = 0

x + yy'  - ky' = 0

y'  ( y - k)  =  -x

y' =     [ x ] / [k - y]

 

The slope of the tangent line to the parabola at any point =   d/dx (x^2) =     2x

 

So....equating these slopes, we have that

 

2x =  x / [ k - y]

2 = 1 / [ k - y]

k - y =    1/2

y = k -1/2      this is the y coordinate of the intersection of the parabola and the circle

 

So.....the difference in the height of the circle's center and the tangent point(s) is just the difference in their  y values

 

So

 

k - (k - 1/2)  =    1/2   unit

 

 

cool cool cool

 Dec 29, 2018
edited by CPhill  Dec 29, 2018
 #3
avatar+800 
-1

Great! It took me a couple minutes to understand, but once I got your method it worked out! Thanks!

CoolStuffYT  Dec 30, 2018
 #4
avatar+103131 
+1

OK, CS.....!!!!

 

[ BTW....this is kind of a neat problem..... ]

 

 

cool cool cool

CPhill  Dec 30, 2018

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