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 \(k\).

FlyEaglesFly 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

CPhill May 7, 2019