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%
