I have the equation not in the center, i.e.
$$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1.$$
But what will be the equation once it is rotated?
Geometry – General Equation of a Rotated Ellipse Not at Origin
analytic geometryconic sectionsgeometry
Related Solutions
Let's start with the parametric equation for a circle centered at the origin with radius 1:
x(t) = cos 2πt
y(t) = sin 2πt
To turn this into an ellipse, we multiply it by a scaling matrix of the form
| a 0 |
| 0 b |
which gives
| a 0 | | x(t) | | a x(t) |
| 0 b | * | y(t) | = | b y(t) |
To rotate this by θ degrees, multiply it by the rotation matrix
| cos θ -sin θ | | a x(t) | |a x(t) cos θ - b y(t) sin θ|
| sin θ cos θ | * | b y(t) | = |a x(t) sin θ + b y(t) cos θ|
So the new parametric equation would be
x'(t) = a x(t) cos θ - b y(t) sin θ = a cos 2πt cos θ - b sin 2πt sin θ
y'(t) = a x(t) sin θ + b y(t) sin θ = a cos 2πt sin θ + b sin 2πt cos θ
You can then translate this to have center (x0, y0) by adding these components to the parametric components.
Hope this helps!
The parameters of the rotation you describe are Tait-Bryan angles. Exactly what rotation they represent depends on several things: the sequence in which you apply the rotations, whether they are intrinsic (body-axis) or extrinsic (fixed-axis) rotations, whether you have a "right-handed" or "left-handed" set of $x,y,z$ axes, and whether a positive rotation angle represents a "right-handed" or "left-handed" rotation around each axis.
A typical setup might be as follows: using a right-handed coordinate system and using a positive angle to represent any right-handed rotation, perform intrinsic rotations first around the $z$-axis through an angle $\gamma$, then around the (rotated) $y$-axis of the body through an angle $\beta$, and finally around the (twice-rotated) $x$-axis of the body through an angle $\alpha$.
Since you also want the center of the body to be offset from the origin of coordinates, after performing the rotations we will translate the entire body by the same translation that takes $(0,0,0)$ to $(x_0,y_0,z_0)$.
If $(x',y',z')$ are the coordinates of a point on the body prior to rotation, and $(x,y,z)$ are the coordinates of the image of that point after rotation and translation, then $$ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = R_z(\gamma)R_y(\beta)R_x(\alpha) \begin{pmatrix} x' \\ y' \\ z' \end{pmatrix} + \begin{pmatrix} x_0 \\ y_0 \\ z_0 \end{pmatrix} $$ where $$ R_x(\alpha) = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos\alpha & -\sin\alpha \\ 0 & \sin\alpha & \cos\alpha \end{pmatrix}, \quad R_y(\beta) = \begin{pmatrix} \cos\beta & 0 & \sin\beta \\ 0 & 1 & 0 \\ -\sin\beta & 0 & \cos\beta \end{pmatrix}, \quad\text{and} $$ $$ R_z(\gamma) = \begin{pmatrix} \cos\gamma & -\sin\gamma & 0\\ \sin\gamma & \cos\gamma & 0 \\ 0 & 0 & 1 \end{pmatrix}. $$ (Yes, that means the rotation matrix for angle $\alpha$ around the $x$-axis is applied first. The use of intrinsic rotations rather than extrinsic rotations causes the sequence of matrix multiplications to be reversed.)
If the original coordinates of all points on a body satisfy the equation $f(x',y',z') = 0$, then in order to find an equation satisfied by the image of that body after the rotation and translation described above, we first write $$ \begin{pmatrix} x' \\ y' \\ z' \end{pmatrix} = R_x(-\alpha) R_y(-\beta) R_z(-\gamma) \begin{pmatrix} x - x_0 \\ y - y_0 \\ z - z_0 \end{pmatrix}. $$ After fully working out the matrix multiplications on the right-hand side, the result is a column vector in which each element is some function of $x$, $y$, $z$, $x_0$, $y_0$, $z_0$, $\alpha$, $\beta$, and $\gamma$. That is, we now have each of the coordinates $x'$, $y'$, and $z'$ of a point of the the unrotated, untranslated object (on the left-hand side of the vector equation) expressed in terms of the coordinates $x$, $y$, and $z$ of the rotated and translated image of the object, along with parameters of the rotation and translation. We can use these equations to substitute for $x'$, $y'$, and $z'$ in the equation $f(x',y',z') = 0$, resulting in a new equation, $g(x,y,z) = 0$.
This is assuming that you perform the rotation and translation exactly as described in the paragraph about "a typical setup". Depending on exactly how you actually perform your rotation and translation, you may have to change the order in which the matrix/vector operations are performed, or you may have to reverse the signs of some angles in the matrices, or both.
Then if you want an equation that determines a rotated and translated ellipsoid, develop the equations for $x'$, $y'$, and $z'$ in terms of $x$, $y$, $z$, and the parameters of the rotation and translation, according to the formulas above or whatever variation of them is appropriate to your particular rotation and translation, and use these equations to substitute for $x'$, $y'$, and $z'$ in $$ f(x',y',z') = \frac{x'^2}{a^2} + \frac{y'^2}{b^2} + \frac{z'^2}{c^2} - 1 = 0. $$ The resulting equation in $x$, $y$, and $z$ is the equation of the rotated, translated ellipsoid.
You could further "simplify" the equation by multiplying out the squared expressions and recombining terms into a more standard polynomial form with terms in $x^2$, $y^2$, $z^2$, $xy$, $yz$, $xz$, $x$, $y$, $z$, and a constant, but I think that would obscure the meaning of the equation (namely, that it describes an ellipsoid) and might not be desirable.
Best Answer
After a lot of mistakes I finally got the correct equation for my problem:-
$$\dfrac {((x-h)\cos(A)+(y-k)\sin(A))^2}{a^2}+\dfrac{((x-h) \sin(A)-(y-k) \cos(A))^2}{b^2}=1,$$
where $h, k$ and $a, b$ are the shifts and semi-axis in the $x$ and $y$ directions respectively and $A$ is the angle measured from $x$ axis.