Answer: \(3\sqrt5\).

From the line

\(x+2y=0,\)

we have

\(x=-2y.\)

Substitute this into the curve equation:

\(\frac4{x^2}+\frac5{4y^2}=1.\)

Since \(x^2=4y^2\),

\(\frac4{4y^2}+\frac5{4y^2}=1.\)

So

\(\frac9{4y^2}=1.\)

Hence

\(4y^2=9,\qquad y=\pm\frac32.\)

When \(y=\frac32\), \(x=-3\). When \(y=-\frac32\), \(x=3\).

The two points are therefore

The distance between them is

Thus