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A circle is tangent to the parabola $y = \frac{x^2}{4}$ at $(2,1),$ and to the $y$-axis. Find the radius of this circle.

 Dec 30, 2020

Best Answer 

 #4
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+3

As follows:

 Dec 30, 2020
 #1
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We can use calculus.  For the parabola y = x^2/4, y' = 2x/4 = x/2.  So for the point (a,a^2/4), the derivative is a/2.  The slope of the normal is -2/a.

 

For the circle, (x - r)^2 + y^2 = r^2.  Then by implicit differentation, 2(x - r) + 2y*2y/dx = 0.  Setting dx = -a and dy = -2, and matching the parameters and solving, we get a = 2 + sqrt(3) and r = 7 - 3*sqrt(3).

 

So the radius of the circle is 7 - 3*sqrt(3).

 Dec 30, 2020
 #2
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Sorry your answer was incorrect, I have no idea about your solution but thanks for the effort

 Dec 30, 2020
 #4
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+3
Best Answer

As follows:

Alan Dec 30, 2020
 #5
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+1

One question only; why is the vertical height from the x-axis to the circle centre \(1+\frac{r}{\sqrt{2}}\) ? Thanks!

Guest Dec 30, 2020
 #6
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+2

Draw a horizontal line through point (2,1) until it is directly below the circle centre. Draw a vertical line from there to the circle centre. You now have a right angled, isosceles triangle with the radius as the hypotenuse (remembering that the hypotenuse here is at 45degrees ). Thus the two legs of this right angled triangle must be of length \(\frac{r}{\sqrt 2}\) by Pythagoras. Add this to the height (1) of the point (2,1).

Alan  Dec 30, 2020
 #7
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+1

Oh!! Thanks that was very helpful!!

Guest Dec 31, 2020

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