What is the general equation of the ellipse that is not in the origin and rotated by an angle?
This Post discusses the formula for an ellipse rotated by an angle. Is a similar formula valid for hyperbola? I think it will be $$\frac{((x−h)\cos A+(y−k)\sin A)^2}{a^2}-\frac{((x−h)\sin A−(y−k)\cos A)^2}{b^2}=1$$
Note: I have no idea how to write an equation using code.If someone could do that, that would be great!
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It’s a straightforward matter to derive the equation yourself. Rotating a point $(x'',y'')$ through an angle $\theta$ about the origin is done via the transformation $$x' = x''\cos\theta-y''\sin\theta \\ y' = x''\sin\theta+y''\cos\theta$$ and a translation of $(x',y')$ by $(h,k)$ is $$x = x'+h \\ y = y'+k.$$ Combine these two transformations and solve for $x''$ and $y''$ in terms of $x$ and $y$, then substitute into the basic equation $\left({x'' \over a}\right)^2-\left({y'' \over b}\right)^2 = 1$.
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The standard equation of an hyperbola in origin is $${x^2\over a^2}-{y^2\over b^2}=1$$We first rotate the hyperbola around the origin and then transport it to some arbitrary point. The rotation matrix is $$\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}$$then by applying it to the standard equation of the hyperbola we obtain $$x'=x\cos\theta-y\sin\theta\\y'=x\sin\theta+y\cos\theta$$which means that$$x=x'\cos\theta+y'\sin\theta\\y=-x'\sin\theta+y'\cos\theta$$so the new equation of the rotated hyperbola is the same you attained whose center is transported to some point $(h,k)$.
For parameterization, notice that the standard form is $$x=a\cosh t\\y=b\sinh t$$so the new parameterization would become $$x=a\cosh t\cos\theta-b\sinh t\sin\theta\\y=a\cosh t\sin\theta+b\sinh t\cos\theta$$
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Also you can draw rotated hyperbola using pair of functions $$y_1=\frac{2}{x+\sqrt{\frac{4}{\tan (b)-\tan (a)}+x^2}}+x \tan (b)$$ and $$y_2=\frac{2}{x-\sqrt{\frac{4}{\tan (b)-\tan (a)}+x^2}}+x \tan (b)$$ connected at points $$\left(\frac{-2}{\sqrt{\tan (a)-\tan (b)}},\frac{-\tan (a)-\tan (b)}{\sqrt{\tan (a)-\tan (b)}}\right)$$ and $$\left(\frac{2}{\sqrt{\tan (a)-\tan (b)}},\frac{\tan (a)+\tan (b)}{\sqrt{\tan (a)-\tan (b)}}\right)$$ where $a$ and $b$ - angles of asymptotes. Sample image.
From @amd, we have that $$(x-h)\sin\theta=x''\sin\theta\cos\theta-y''\sin^2\theta\tag{1}$$$$(y-k)\cos\theta=x''\sin\theta\cos\theta+y''\cos^2\theta\tag{2}$$ Now $(2)-(1)$ gives $$y''=(y-k)\cos\theta-(x-h)\sin\theta\tag{3}$$ using the identity $\sin^2\theta+\cos^2\theta=1$.
Hence putting $(3)$ into $y-k=x''\sin\theta+y''\cos\theta$, we get $$x''=\frac{y-k-(y-k)\cos^2\theta+(x-h)\sin\theta\cos\theta}{\sin\theta}=(y-k)\sin\theta+(x-h)\cos\theta\tag{4}$$ since $1-\cos^2\theta=\sin^2\theta$.
Combining $(3)$ and $(4)$, we get $$\left(\frac{(y-k)\sin\theta+(x-h)\cos\theta}a\right)^2-\left(\frac{(y-k)\cos\theta-(x-h)\sin\theta}b\right)^2=1\tag{5}$$