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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

CoolStuffYT Dec 29, 2018

#1**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?

ElectricPavlov Dec 29, 2018

#5**-1 **

Sorry for answering late...

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

CoolStuffYT
Jan 1, 2019

#2**+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

CPhill Dec 29, 2018

#3**-1 **

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

CoolStuffYT
Dec 30, 2018