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?

**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.

**Attribution***Source : Link , Question Author : andikat dennis , Answer Author :
4 revs, 4 users 44%*