Find the area of the region containing the points inside the ellipse
$2x^2+y^2-2xy-4x=0$
but outside the ellipse
$2x^2+y^2+2xy-4x=0$.
I was able to find the required area using integration. But it is required to find the area without using integration. Is there a property of ellipse that i am missing.
Your problem amounts at finding the area $A_\text{elliptical segment}$ of that segment of ellipse $e$ of equation $$ 2x^2+y^2+2xy−4x=0 $$ (blue in figure below) lying above the $x$-axis. The area of the whole region mentioned in the question is twice that area.
From the equation one readily finds that $e$ intersects the $x$-axis at $A=(0,0)$ and $B=(2,0)$. Moreover, for $x=0$ and $x=4$ we get a single value of $y$, thus $e$ has vertical tangents at $A$ and $(4,-4)$, and its center is their midpoint $O=(2,-2)$. The line through $O$ parallel to $x$-axis intersects $e$ at $H=(2-\sqrt2,-2)$, $I=(2+\sqrt2,-2)$.
If $M=(1,0)$ is the midpoint of $AB$, then line $OM$ intersects $e$ at points $E$, $F$ with horizontal tangent. Moreover, $OM$ meets at $T=(0,2)$ both tangents at $A$ and $B$, and we have: $$ OM:OE=OE:OT, $$ which projected onto $y$-axis gives $$ 4:(y_E+2)=(y_E+2):2, \quad\text{that is:}\quad y_E=2\sqrt2-2. $$ Let's now shear $e$ about the $x$-axis, to transform it to an ellipse $e'$ whose axes are parallel to cartesian axes (red in figure below). Note that the semi-major axis of $e'$ has a length $$a'=y_E-y_O=2\sqrt2,$$ while the semi-minor axis has a length $$b'=x_I-x_O=\sqrt2.$$ Shearing preserves area, hence the area we want to compute is the same as that of the part of ellipse $e'$ lying above the $x$-axis.
Finally, we can stretch $e'$ perpendicular to $x$-axis by a factor $b'/a'=1/2$, to obtain a circle (green in the figure) with radius $b'$. The area of the circular segment with base $AB$ lying above the $x$-axis can be easily computed as: $$ A_\text{circular segment}={\pi\over2}-1. $$ To get the area of the elliptical segment we need only un-stretch $A_\text{circular segment}$ multiplying it by a factor of $2$: $$ A_\text{elliptical segment}={\pi}-2. $$