Transforming coordinates system to obtain an ellipse in the standard form

Question

Let $E$ be the ellipse $x^2+xy +y^2 = 1$. I would like to obtain the area of $E$ with the formula $ab\pi$. I am not able to transform the equation in the form $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2} = 1$. The textbook I am following uses a transformation to describe $E$ as $$\dfrac{X^2}{\left(\dfrac{1}{\sqrt{3}}\right)^2}+\dfrac{Y^2}{1^2}=1$$

I do not understand what intermediate steps lead to this transformation.

EDIT : As some were curious, this is the original question:

Use Stoke's Theorem to to evaluate $\oint_C\vec{F}\cdot\vec{dr}$. Assume C is oriented counterclockwise as viewed from above.

$$\vec{F} = z\vec{i}+x\vec{j} + y\vec{k}$$

$C$ is the curve of intersection of the plane $x + y + z = 0$ and the sphere $x^2 + y^2 + z^2 = 1$.

Hint : recall that the area of an ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2} = 1$ is $ab\pi$

The textbook solution : 3 * Area of R = $3 \cdot \dfrac{\pi}{\sqrt(3)}$=$\sqrt3\pi$

Answer 1

Let $X=\dfrac x2$ and $Y=\dfrac x2 + y$.

Then $\dfrac{X^2}{\left(\dfrac{1}{\sqrt{3}}\right)^2}+\dfrac{Y^2}{1^2}=3X^2+Y^2=\dfrac34x^2+\dfrac14x^2+xy+y^2=1$.

Addendum:

The above explains the intermediate steps that led to the transformation, as the original post requested, but it should be noted, as alluded to in the comments, that the transformation

$\begin{pmatrix}X\\Y\end{pmatrix}=\begin{pmatrix}\frac12 & 0 \\ \frac12 &1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$ has determinant $\frac12$, so this transformation halves area, and that must be taken into account when addressing the underlying question about the area of the ellipse.

Answer 2

The better solution of your original problem is indeed to apply Stokes's Theorem on the original curve. Take the filled in disk bounded by the circle $C$ in the plane $x+y+z=0$. When you compute $\text{curl}\,\vec F\cdot\vec n$, you will get a constant. Then multiply this constant by the area of the circle (easy enough, since it's a circle of radius $1$).

EDIT: The textbook answer is correct, but their “derivation” is highly suspect.

Answer 3

This ellipse contains the six points $(\pm1,0)$, $(0,\pm1)$, $(-1,1)$, and $(1,-1)$. It’s symmetric about both diagonals through the origin.

Thus the diagonal $y=-x$ is the major axis, and $y=x$ is the minor axis; the endpoints of the major axis are $(-1,1)$ and $(1,-1)$, so that the semimajor axis is of length $\sqrt2$, while the endpoints of the minor axis are $\pm(\frac1{\sqrt3},\frac1{\sqrt3})$, the semiminor axis thus being of length $\frac{\sqrt2}{\sqrt3}$.

Thus I make the area to be $$\frac{2\pi}{\sqrt3}\,,$$ while your text seems to be claiming that the area is $\pi/\sqrt3$.