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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
+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   May 7, 2019
#2
-1

thank you that really helps

FlyEaglesFly  May 7, 2019