What is the general equation of the ellipse that is not in the origin and rotated by an angle? 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?
 A: 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.
A: If you came here looking for how
$$c_0x^2 + c_1y^2 + c_2xy + c_3x + c_4y + c_5 = 0$$
relates to $h, k, a, b, A$ its the following:
\begin{eqnarray}
c_0&=&\frac{\cos^2(A)}{a^2} + \frac{\sin^2(A)}{b^2}\\
c_1&=&\frac{\sin^2(A)}{a^2} + \frac{\cos^2(A)}{b^2}\\
c_2&=&\frac{\sin(2A)}{a^2} - \frac{\sin(2A)}{b^2}\\
c_3&=&-\frac{2 h \cos^2(A)}{a^2} - \frac{k\sin(2A)}{a^2} - \frac{2 h \sin^2(A)}{b^2} + \frac{k \sin(2A)}{b^2}\\
c_4&=&-\frac{h \sin(2A)}{a^2} - \frac{2k\sin^2(A)}{a^2} + \frac{h \sin(2A)}{b^2} - \frac{2k \cos^2(A)}{b^2}\\
\end{eqnarray}
$$c_5 = \frac{h^2 \cos^2(A)}{a^2} + \frac{h k \sin(2A)}{a^2} + \frac{k^2 \sin^2(A)}{a^2}  + \frac{h^2 \sin^2(A)}{b^2} - \frac{ h k \sin(2A)}{b^2} + \frac{k^2 \cos^2(A)}{b^2} - 1
$$
A: The equation you gave can be converted to the parametric form:
$$
x = h + a\cos\theta \quad ; \quad y = k + b\sin\theta
$$
If we let $\mathbf x_0 = (h,k)$ denote the center, then this can also be written as
$$
\mathbf x = \mathbf x_0 + (a\cos\theta)\mathbf e_1 +  (b\sin\theta)\mathbf e_2
$$
where $\mathbf e_1 = (1,0)$ and $\mathbf e_2 = (0,1)$.
To rotate this curve, choose a pair of mutually orthogonal unit vectors $\mathbf u$ and $\mathbf v$, and then
$$
\mathbf x = \mathbf x_0  + (a\cos\theta)\mathbf u +  (b\sin\theta)\mathbf v
$$
One way to define the $\mathbf u$ and $\mathbf v$ is:
$$
\mathbf u = (\cos\alpha, \sin\alpha) \quad ; \quad \mathbf v = (-\sin\alpha, \cos\alpha)
$$
This will give you an ellipse that's rotated by an angle $\alpha$, with center still at the point $\mathbf x_0 = (h,k)$.
If you prefer an implicit equation, rather than parametric ones, then any rotated ellipse (or, indeed, any rotated conic section curve) can be represented by a general second-degree equation of the form
$$
ax^2 + by^2 + cxy + dx + ey + f = 0
$$
The problem with this, though, is that the geometric meaning of the coefficients $a$, $b$, $c$, $d$, $e$, $f$ is not very clear.
There are further details on this page.
Addition. Borrowing from rschwieb's solution ...
Since you seem to want a single implicit equation, proceed as follows. Let  $c = \sqrt{a^2 - b^2}$. Then the foci of the rotated ellipse are at $\mathbf x_0 + c \mathbf u$ and $\mathbf x_0 - c \mathbf u$. Using the "pins and string" definition of an ellipse, which is described here, its equation is
$$
\Vert\mathbf x - (\mathbf x_0 + c \mathbf u)\Vert + 
\Vert\mathbf x - (\mathbf x_0 - c \mathbf u)\Vert = \text{constant}
$$
This is equivalent to the one given by rschwieb. If you plug $\mathbf u = (\cos\alpha, \sin\alpha)$ into this, and expand everything, you'll get a single implicit equation.
The details are messy (which is probably why no-one wants to actually write everything out for you).
A: As stated, using the definition for center of an ellipse as the intersection of its axes of symmetry, your equation for an ellipse is centered at $(h,k)$, but it is not rotated, i.e. the axes of symmetry are parallel to the x and y axes. 
If this were not true, you would have a cross-product term involving $x \times y$. If you had such a term, you could calculate the counterclockwise rotation angle $\alpha$ required in order to eliminate the cross-product term (and thereby make the axes of symmetry parallel to the x and y axes).
One way is to use the formula $$\cot 2\alpha = \frac{A - C}{B},$$ where $\alpha$ is the counterclockwise rotation angle, $A$ is the coefficient of $x^2$, $B$ the coefficient of the cross-product term $x \times y$, and $C$ is the coefficient of $y^2$. 
In order to apply the rotation once you know $\alpha$, you can find new coordinates $x', y'$ in terms of $x, y$ via $x' = x \cos \alpha - y \sin \alpha$ and $y' = x \sin \alpha + y \cos \alpha$.
Source: Calculus and Analytic Geometry, by George Thomas (paraphrased).
A: Another option is to use the geometric definition of an ellipse as the set of points whose sum distance to the foci is constant.
If the foci are at $(a,b)$ and $(a',b')$, and the sum distance is $C$, you get:
$$\sqrt{(x-a)^2+(y-b)^2}+\sqrt{(x-a')^2+(y-b')^2}=C$$
A: If you want the center to be $(h,k)$


*

*first apply a general rotation of coordinates transformation to 
$$\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$$
to rotate the axes to whatever angle you desire.

*then translate the center to $(h,k)$ by replacing the new $x$ and $y$ by
$(x-h)$ and $(y-k)$.
