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
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?
Sorry for answering late...
CPhill already got the answer. If you could look at his, you may understand :)
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
Great! It took me a couple minutes to understand, but once I got your method it worked out! Thanks!