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From the above picture we can prove easily $ CF = XF$ because CFX is a isosceles triangle. Now if we move the point X more near near P, it will give the same results(I mean we will get again $ CF = XF$) but by drawing the figure on paper, I have measured that XF= CF are not equal always. What is wrong here?

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  • $\begingroup$ a-levelphysicstutor.com/images/optics/ccmrrs-r-2f.jpg $\endgroup$ – Physics_guy May 24 '13 at 20:33
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    $\begingroup$ What is the nature of the particular curve? I would have guessed that it’s supposed to be a parabola. What gives you the idea that the curve is accurately drawn? $\endgroup$ – Lubin May 24 '13 at 21:19
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    $\begingroup$ if you move the point X more near near P, it will give the same results iff CF=FP.. $\endgroup$ – mle May 24 '13 at 21:28
  • $\begingroup$ why was this migrated to MSE? This is obviously a question on geometrical optics, it belongs here (geometry) as much as it belongs there (optics). $\endgroup$ – user31280 May 24 '13 at 21:31
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    $\begingroup$ Here you go. $\endgroup$ – newbie May 25 '13 at 0:39
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The geometry in the diagram is true no matter what the figure of the mirror is; that is, $\triangle XFC$ is isosceles. What is not true is that both $C$ and $F$ are independent of $X$. In what follows, $\kappa$ is the curvature of the curve at $P$.

If we wish to fix $F$ independent of $X$, then the curve needs to be a parabola. However, in that case, $C$ is dependent on $X$.

$\hspace{2cm}$enter image description here

If we wish to fix $C$ independent of $X$, then the curve needs to be a circle. However, in that case, $F$ is dependent on $X$.

$\hspace{2cm}$enter image description here

Note that as $r\to0$, $PC\to2PF$.

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