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An ellipse is defined by the equation x^2 /9 + y^2 /4 = 1. Compute the radius of the largest circle that is internally tangent to the ellipse at (3, 0) and intersects the ellipse only at (3, 0).

 Jan 11, 2022
 #1
avatar+2401 
+3

Let r be the radius of the circle. 

The circle must have a center of (3 - r, 0).

So our formula is (x - 3 + r)^2 + y^2 = r^2

9(x - 3 + r)^2 + 9y^2 = 9r^2

x^2 /9 + y^2 /4 = 1

4x^2 + 9y^2 = 36

Subtracting the two formulas. 

9(x - 3 + r)^2 - 4x^2 = 9r^2 - 36

5x^2 + (18r - 54)x + (117 - 54r) = 0

There is one solution when the discriminant is 0. 

(18 - 54r)^2 - 4(5)(117 - 54r) = 0

r = 2(2 + sqrt(130))/27 (note that r can't be negative so it's not the other option)

 

Hope this was helpful. 

=^._.^=

 Jan 12, 2022
 #2
avatar+117443 
+3

Thanks Catmg,

 

That looks like a bit of a challenge. :)

I just thought I'd take a graphical look at your result.  Looks reasonable :)

 

https://www.desmos.com/calculator/zbydksn1f1

 

 Jan 13, 2022
 #3
avatar+2401 
+1

Thank you, it's always nice to see the problem in a graph. :))

 

=^._.^=

catmg  Jan 19, 2022
 #4
avatar+117443 
+1

You did all the work catmg :)

Melody  Jan 19, 2022

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