Let's assume we have the foci of a hyperbola as $(6,1)$ and $(15,3)$ and that the length of the transverse axis is $2b$ units.
Now since the foci and transverse axis are specified,the hyberbola is uniquely determined. Again we know the property that the difference of focal distances is equal to the transverse axis. But how do we apply that property here?
I mean if our moving point is $(x,y)$,which of $\sqrt{(x-6)^2+(y-1)^2}-\sqrt{(x-15)^2+(y-3)^2}=2b$
Or $\sqrt{(x-15)^2+(y-3)^2}-\sqrt{(x-6)^2+(y-1)^2}=2b$ is correct?
Surely both will yeild different equations of the hyperbola,but only one can be right since the hyperbola is unique. Then how do we understand which equation is the correct one?
((x-15)^2+(y-3)^2-((x-6)^2+(y-1)^2)-4*b^2)^2-16*b^2*((x-6)^2+(y-1)^2)or the other((x-6)^2+(y-1)^2-((x-15)^2+(y-3)^2)-4*b^2)^2-16*b^2*((x-15)^2+(y-3)^2)they become the same equation, but if you don't expand all the way you get the focus directrix versions of the equation $$(\frac{-18x-4y-4b^2+197}{4b})^2=\frac{85}{4b^2}\frac{(-\frac92x-y-b^2+197/4)^2}{\frac{85}{4}}=(x-6)^2+(y-1)^2$$ $$(\frac{18x+4y-4b^2-197}{4b})^2=\frac{85}{4b^2}\frac{(\frac92x+y-b^2-197/4)^2}{\frac{85}{4}}=(x-15)^2+(y-3)^2$$ revealing the squared eccentricity to be $\frac{85}{4b^2}.$ – Jan-Magnus Økland Jan 20 '23 at 10:02