Let $ABCD$ be a quadrilateral circumscribing the ellipse $$\frac{x^2}{25}+\frac{y^2}{16}=1.$$ Let $S$ be one of its focii, then what is the sum of the angles $\angle{ASB}$ and $\angle{CSD}$?
My try: I have done this for the extreme case of quadrilateral, taking quadrilateral to be rectangle and got the answer which is true for every quadrilateral, but how to do it for general quadrilateral?
On
Here, I am not very sure about where ABCD are. I am assuming A is at the top left, B at the top right, C at the bottom right, and D at the bottom left. If this interpretation of the question is wrong, please correct me.
We know that the ellipse is "lying down", with a width of $2*5=10$ and a height of $2*4=8$. Then $\angle ASB=180^o-tan^{-1}(\frac{4}{2})-tan^{-1}(\frac{4}{8})=90^o$.
I am using here $\triangle ASE$, where E is the midpoint of AD, for $tan^{-1}(\frac{4}{2})$.
I am using $\triangle BSF$, where F is the midpoint of B and C, for $tan^{-1}(\frac{4}{8})$.
Multiplying this($90^o$) by 2, since $\angle ASB=\angle CSD$, we get $180^o$. Over here, we conclude that whatever the interpretation, it is always $180^o$.
The intersection point of two tangents to an ellipse lies on the bisector of the angle formed by joining a focus to the points of tangency (for a proof see HERE). Hence $$\angle{ASB}=\angle{ASD}\quad\mbox{and}\quad\angle{CSD}=\angle{CSB}.$$ Since $$\angle{ASB}+\angle{ASD}+\angle{CSD}+\angle{CSB}=360^{\circ}$$ we may conclude that $\angle{ASB} + \angle{CSD}=180^{\circ}$. Note that this property holds for ANY ellipse.