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A beam of light travelling between the points (0, $y_0$) and $(x_1, y_1)$ in the (x, y) plane will travel along the curve y(x) where y' = dy/dx.

I have worked out that the path y(x) must satisfy the equation

$$y\sqrt{1+y'^2}=\frac{1}{c}$$

the beam of light is travelling within the infinitely wide strip between y=$y_0$ and y=$2y_0$ where $y_0$>0

By substitution or otherwise, I must verify that the circular path satisfies my previous equation.

$y(x)^2+ (x-x_0)^2=\frac{1}{c^2}$

I have tried making the equations equal each other but that has come up blank I think I have to find y'(x) when I try to differentiate y(x) all I get is y'(x)=-1 and that doesn't help very much either. all help would be greatly appreciated as this question has me really stumped. feel free to adapt tags as you see fit I wasn't sure what to put for this question

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Square the first equation and set it equal to the second $$y^2+ (x-x_0)^2=y^2(1+(y')^2 ) \\[9pt] \iff(x-x_0)^2 = (yy')^2$$ Now implicitly differentiate the second equation

$$2yy' +2(x-x_0) = 0 \iff yy'=-(x-x_0) $$

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  • $\begingroup$ $y$ is the name of a function of $x$, $y(x)$ and $y$ mean the same thing. $\endgroup$
    – WW1
    Apr 27 at 2:44
  • $\begingroup$ thanks for clearing that up for me $\endgroup$
    – user833726
    Apr 27 at 2:45

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