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Let \(F_1\) and \(F_2\) be the foci of the ellipse \(kx^2 + y^2 = 1,\) where \(k > 1\) is a constant. Suppose that there is a circle which passes through \(F_1\) and \(F_2\) and which lies tangent to the ellipse at two points on the \(x\)-axis. Compute \(k\).

 May 7, 2019
 #1
avatar+106533 
+1

The focal points will lie on the y axis.....and the circle will be centered at the origin

 

kx^2 + y^2  = 1    we can write

 

   x^2                 y^2

  _____   +       _____  =    1

 (1/√k)^2              1

 

The  focal distance will be  =  √[ 1 - (1/√k)^2]  =  √[ 1 - 1/k ]

The radius of the circle will be the square of this

 

So  ... since the circle is tangent to the ellipse at two points on the x axis.....the point  (  √ (1 - 1/k) , 0 )  will be on the ellipse...so 

 

[√ (1 - 1/k)]^2                0

__________  +         ___    =     1         simplifying, we have

    (1/ √k)^2                  1

 

1 - 1/k  =  1/k

1 = 2/k

k = 2

 

Here's a graph :  https://www.desmos.com/calculator/nlcn74kcb7

 

 

 

cool cool cool

 May 7, 2019
 #2
avatar+166 
0

thank you that really helps

FlyEaglesFly  May 7, 2019

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