I want to find the area under the curve:
$$(2018x+2y-1)^2+ (2018x-3y+2)^2 =1$$
What I tried is:
Firstly, I substituted $2018x=X$ and kept $y=Y$, as this would scale my area by $2018$, so in the end, we have to divide the answer by $2018$. Further, I found that it was not possible to show $y=f(x)$ so I tried converting it into a polar system as $r=f(\theta)$. Therefore I substituted $X= r \cos(\theta)$ and $Y=r \sin(\theta)$. After substituting these, I got a quadratic in $r$ as:
$$r^2(2+11\sin^2 \theta - \sin 2 \theta) +r(2 \cos \theta-16 \sin \theta)+4=0$$
But I don't think this could be simplified any further. I think I am going in the wrong way, so please comment if I am. Also, the answer was $\frac{\pi}{10090}$ if anyone solves it.
Since you are looking for the area bounded by the curve, then note that the area does not change if you move the centre. Therefore, you can find the area bounded by:
$$(x+2y)^2+(x-3y)^2 = 1$$
and then divide by $2018$.
Now substitute $\cos \theta = x +2y, \sin \theta = x - 3y$ due to the Pythagorean identity. Then solve for $x$ and $y$, where we get $y = \frac{\cos \theta - \sin \theta}{5}, x = \frac{3 \cos \theta + 2 \sin \theta}{5}$.
The extrema (turning points) of the ellipse can be found by finding the maximum and minimum points of $\frac{\cos \theta - \sin \theta}{5}$ and $\frac{3 \cos \theta + 2 \sin \theta}{5}$. You can do this by expressing the equations in the form $a \cos( \theta + \alpha)$. Since the ellipse is a linear transformation centred at the origin, the extrema are actually symmetric around the origin.
Once you have the four turning points, find the distance between each pair. Then use the fact that the area of an ellipse is $\pi a b$.