In this Desmos graph, it shows you how to rotate a parabola. Is there an equation similar to the one in the Desmos graph that allows a hyperbola to be rotated around an arbitrary point on a Cartesian plane by a certain amount of degrees? Presumably without matrices. I tried this equation from this Stack Exchange answer: $$\left(\frac{(y-k)\sin\theta-(x-h)\cos\theta}a\right)^2-\left(\frac{(y-k)\cos\theta-(x-h)\sin\theta}b\right)^2=1$$ But, in Desmos, it kept saying there are too many variables and doesn't know what to do with it.
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Your difficulty seems to be caused by Desmos rather than the equation itself. I suggest you ask at Stack Overflow or Computational Science SE. (By the way, your first link is not working.) – sammy gerbil Apr 02 '20 at 04:35
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@sammygerbil Sorry. I meant to copy paste the desmos link into the hyperlink but I accidentally copied the equation. – Star Man Apr 02 '20 at 16:28
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If the point $(x_1,y_1)$ is rotated around $(x_0,y_0)$ through an angle $\theta$ to give the point $(x_1',y_1')$, $$x_1'-x_0=(x_1-x_0)\cos \theta-(y_1-y_0)\sin \theta$$ $$y_1'-y_0=(x_1-x_0)\sin \theta+(y_1-y_0)\cos \theta$$ and thus $$x_1-x_0=(x_1'-x_0)\cos \theta+(y_1'-y_0)\sin \theta$$ $$y_1-y_0=-(x_1'-x_0)\sin \theta+(y_1'-y_0)\cos \theta$$ which gives $$x_1=(x_1'-x_0)\cos \theta+(y_1'-y_0)\sin \theta+x_0$$ $$y_1=-(x_1'-x_0)\sin \theta+(y_1'-y_0)\cos \theta+y_0$$ Take the equation of the hyperbola and relace $x$ by $$(x-x_0)\cos \theta+(y-y_0)\sin \theta+x_0$$ and $y$ by $$-(x-x_0)\sin \theta+(y-y_0)\cos \theta+y_0.$$
P. Lawrence
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