Problem about an ellipse and its eccentricity If the tangent at a point (a cosθ,b sinθ) on the ellipse meets the auxiliary circle in two points, the chord joining them subtends a right angle at the centre, then the eccentricity of the ellipse is given by which of these?
A)$(1+(cosθ)^2)^{\frac{-1}{2}}$
B) $1+(sinθ)^2$
C)$(1+(sinθ)^2)^{\frac{-1}{2}}$
D) 1+$(cosθ)^2$
I know that eccentricity of an ellipse is equal to R/r-1,where R and r are its diameters. If an ellipse meets an auxiliary circle at two points, I've also tried to use the Power of the Point theorem and the properties of the Radical Axis. But what next? How are all these trigonometric functions tied to the ellipse? Thanks!
 A: This is not the answer you want, but it must be C.  Suppose a is the semi major axis length.  For $\theta=\pi/2$, the two points of intersection with the circle are $(\pm\sqrt{a^2-b^2},b)$.  Your condition then says the dot product of these two points is 0.  From this it follows easily the eccentricity is $1/\sqrt{2}$.  Conversely, for this eccentricity,
A: 
Given ellipse E with equation $b^2x^2+a^2y^2=a^2b^2$ with a>b and auxiliary circle C with equation $x^2+y^2=a^2$, let $P=(x_1,y_1)$ be any point on the ellipse.  Let Q and R be two points of intersection of this tangent with C, $F_1=(-c.0)$ and $F_2=(c,0)$ the foci of E $(c^2=a^2-b^2)$, and O the origin.  Then $OQ\parallel F_1P$ and $OR\parallel F_2P$.
The only proof I have of this result is a rather lengthy analytic geometry proof; if you're interested, I can post it.

Same setup as the first diagram, but here angle ROQ is a right angle.
Triangle $F_1F_2P$ is a right triangle since $F_1P\parallel OQ$ and $F_2P\parallel OR$.  From $|PF_1|^2+|PF_2|^2=2a$, easily $|F_1P|=a+ex_1$ and $|F_2P|=a-ex_2$.  So $(a+ex_1)^2+(a-ex_1)^2=|F_1F_2|^2=4c^2$.  So $a^2+e^2x_1^2=2c^2$ or $a^2(1+e^2\cos^2(\theta))=2c^2$; i.e. $1+e^2(1-\sin^2(\theta)=2e^2$,  $e^2(1+\sin^2(\theta))=1$ and finally 
$$e={1\over\sqrt{1+\sin^2(\theta)}}$$ since $e>0$.
Note the result shows that given the ellipse, there are exactly four such points P.  These are the symmetric points about the axes.  One such is the point $P^{\prime}$, the reflection of P about the x-axis.
P.S. I don't know if it's a beautiful problem, but I enjoyed working out the answer.
