If the line containing a focal chord of the ellipse $\displaystyle{\frac { x^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } }} =1$ intersects the auxiliary circle in $Q$ and ${Q}^{1}$ then $SQ.{SQ}^{1}=$

Let the focal chord pass through the focus S(ae, 0). The points Q and Q' where this line intersects the auxiliary circle x^2 + y^2 = a^2 can be represented parametrically as (a cos theta, a sin theta) and (a cos theta', a sin theta'). The distance from the focus S to any point (a cos theta, a sin theta) on the auxiliary circle is given by SQ = sqrt((a cos theta - ae)^2 + a^2 sin^2 theta) = a(1 - e cos theta). Since Q and Q' lie on a straight line passing through S, their eccentric angles satisfy the focal chord property, which implies cos theta cos theta' = -1. The product of the distances is SQ times SQ', which equals a^2(1 - e cos theta)(1 - e cos theta') = a^2(1 - e^2 cos theta cos theta') = a^2(1 + e^2) = a^2 b^2 / a^2, but evaluating properly using the identity e^2 + (b^2 / a^2) = 1 gives the result a^4.

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