I asked the purely mathematical question over herehere, and received the most complete answer. While I thought there would be a simple trick to seeing the hyperbolic relationship, it looks like you just have to go through the tedious algebra to have it pop out. User JJacquelin found that it can be rearranged to the form $$\frac{z^2}{(\frac{(2n+1)\pi}{k})^2}-\frac{y^2}{L^2-(\frac{(2n+1)\pi}{k})^2}=\frac{1}{4}+\frac{a^2}{L^2-(\frac{(2n+1)\pi}{k})^2}$$ To show how this looks, I set $L = 20 \space \text{cm}$, $a= 2 \space \text{cm}$, and an infrared light source of $\lambda = 1000 \space \text{nm}$. Plotting a modest amount of these produced the following image: 
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
I asked the purely mathematical question over here, and received the most complete answer. While I thought there would be a simple trick to seeing the hyperbolic relationship, it looks like you just have to go through the tedious algebra to have it pop out. User JJacquelin found that it can be rearranged to the form $$\frac{z^2}{(\frac{(2n+1)\pi}{k})^2}-\frac{y^2}{L^2-(\frac{(2n+1)\pi}{k})^2}=\frac{1}{4}+\frac{a^2}{L^2-(\frac{(2n+1)\pi}{k})^2}$$ To show how this looks, I set $L = 20 \space \text{cm}$, $a= 2 \space \text{cm}$, and an infrared light source of $\lambda = 1000 \space \text{nm}$. Plotting a modest amount of these produced the following image:
I asked the purely mathematical question over here, and received the most complete answer. While I thought there would be a simple trick to seeing the hyperbolic relationship, it looks like you just have to go through the tedious algebra to have it pop out. User JJacquelin found that it can be rearranged to the form $$\frac{z^2}{(\frac{(2n+1)\pi}{k})^2}-\frac{y^2}{L^2-(\frac{(2n+1)\pi}{k})^2}=\frac{1}{4}+\frac{a^2}{L^2-(\frac{(2n+1)\pi}{k})^2}$$ To show how this looks, I set $L = 20 \space \text{cm}$, $a= 2 \space \text{cm}$, and an infrared light source of $\lambda = 1000 \space \text{nm}$. Plotting a modest amount of these produced the following image:
I asked the purely mathematical question over here, and received the most complete answer. While I thought there would be a simple trick to seeing the hyperbolic relationship, it looks like you just have to go through the tedious algebra to have it pop out. User JJacquelin found that it can be rearranged to the form $$\frac{z^2}{(\frac{(2n+1)\pi}{k})^2}-\frac{y^2}{L^2-(\frac{(2n+1)\pi}{k})^2}=\frac{1}{4}+\frac{a^2}{L^2-(\frac{(2n+1)\pi}{k})^2}$$ To show how this looks, I set $L = 20 \space \text{cm}$, $a= 2 \space \text{cm}$, and an infrared light source of $\lambda = 1000 \space \text{nm}$. Plotting a modest amount of these produced the following image: