I am told that the image created by a parabolic mirror is the same size as the object when it is on the center of curvature, which is twice away from the focus of the parabola. I tried to prove this in the following way:

Given a parabola with focus $(0,a)$, we obtain the equation $y=x^2/4a$. Now, the center of curvature is at $(0,2a)$, and hence an object of length $l$ must have its highest point at $(\pm l,2a)$. Let's choose $(l,2a)$ as the highest point, in which case the image should have a lowest point $(-l,2a)$.

Now, the standard procedure to find the image is to draw 2 lines from $(l,2a)$; one that intersects the focus (0,a) and another that is parallel to the principal axis of the parabola, as shown by the figure below.

[![Parabola](https://i.stack.imgur.com/zdLzP.png)](https://i.stack.imgur.com/zdLzP.png)

The former creates a line that goes through the points $(l,2a)$ and $(0,a)$ and has the form

$$y=\frac{a}{l}x+a$$

But we see that this doesn't intersect the parabola $y=x^2/4a$ at some point with $x$ coordinate $-l$, leading me to believe that the image isn't actually the same size.

The figure corroborates this geometrically as well.

**Is the image of an object at the center of curvature of a parabolic mirror actually the same size as the object itself?**