How to show that the angle at intersections of concentric ellipses with a radial line is constant? I'm looking for a proof that the angle at intersections of concentric ellipses with a radial line remains constant as you move from one concentric ellipse to another.
For example, a series of concentric ellipses of different sizes are centered on a point of origin.
The radial lines from the point of origin outwards that trace the semi-major and semi-minor axis of the concentric ellipses only form right-angles at each intersection with the concentric ellipses.
How may one show that the other radial lines forming non-perpendicular intersections with the concentric circles form a constant angle with respect to the same radial line?
 A: If someone is not convinced by dilation argument, you can also prove the fact in the following way:
The equation of ellipse is:
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ 
So the slope of the tangent is:
$$\frac{x}{a^2}+\frac{yy'}{b^2}=0$$
$$y'=-\frac{b^2x}{a^2y}\tag{1}$$
Now draw two ellipses with semi-axis $(a_1,b_1)$ and $(a_2,b_2)$. For concentric ellipses (with the same center and eccentricity):
$$\frac{b_1}{a_1}=\frac{b_2}{a_2}\tag{2}$$ 
Draw a line $p$ passing through origin intersecting ellipses in points $(x_1,y_1)$ and $(x_2, y_2)$. Obviously:
$$\frac{x_1}{y_1}=\frac{x_2}{y_2}\tag{3}$$
By combining (1), (2) and (3):
$$y_1'=-\frac{b_1^2x_1}{a_1^2y_1}=-\frac{b_2^2x_2}{a_2^2y_2}=y_2'$$
So both tangents have the same slopes which means that the tangents are parallel and therefore their angles with respect to line $p$ must be equal.
EDIT: I have assumed that concentric ellipses must have the same center and the same eccentricity. Eccentricity of an ellipse is defined with:
$$e=\sqrt{1-\frac{b^2}{a^2}}$$ 
So two ellipses with the same center are concentric only if (2) holds.
