jmc

The two vertical lines have equations of the form $x=m$ and $x=M,$ where $m$ and $M$ are the least and greatest possible $x-$coordinates for a point on the ellipse. Similarly, the horizontal lines have equations of the form $y=n$ and $y=N,$ where $n$ and $N$ are the least and greatest possible $y-$coordinates for a point on the ellipse. Therefore, we want to find the range of possible $x-$ and $y-$coordinates over all points on the ellipse.

Subtracting $5$ from both sides, we can write the equation of the ellipse as a quadratic with $x$ as the variable: $$x^2-(2y)x+(3y^2-5)=0.$$For a point $(x,y)$ to lie on the ellipse, this equation must have a real solution for $x.$ Therefore, the discriminant of the quadratic must be nonnegative: $$(2y{)}^2-4(3y^2-5)\ge 0,$$or $-8y^2+20\ge 0.$ Solving for $y$ gives $-\frac{\sqrt{10}}{2}\le y\le \frac{\sqrt{10}}{2}.$ Therefore, the equations of the two horizontal lines are $y=-\frac{\sqrt{10}}{2}$ and $y=\frac{\sqrt{10}}{2}.$

We can do the same, with the roles of the variables reversed, to find all possible values for $x.$ We write the equation of the ellipse as a quadratic in $y$, giving $$3y^2-(2x)y+(x^2-5)=0.$$The discriminant of this equation must be nonnegative, so we have $$(2x{)}^2-4\cdot 3\cdot (x^2-5)\ge 0,$$or $-8x^2+60\ge 0.$ Solving for $x$ gives $-\frac{\sqrt{30}}{2}\le x\le \frac{\sqrt{30}}{2}.$ Therefore, the equations of the two vertical lines are $x=-\frac{\sqrt{30}}{2}$ and $x=\frac{\sqrt{30}}{2}.$

It follows that the side lengths of the rectangle are $2\cdot \frac{\sqrt{10}}{2}=\sqrt{10}$ and $2\cdot \frac{\sqrt{30}}{2}=\sqrt{30},$ so the area of the rectangle is $$\sqrt{10}\cdot \sqrt{30}=\boxed{10\sqrt{3}}.$$