+170 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 \(k\).
+129899 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
