Let e1 be the eccentricity of the hyperbola $\frac{x^2}{16}-<!--\; -\; -->\frac{y^2}{9}=1$ and e₂ be the eccentricity of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,a>b$, which passes through the foci of the hyperbola. If e₁e₂ = 1, then the length of the cord of the ellipse parallel to the x-axis and passes through (0, 2) is:

The correct answer is:
$\frac{10\sqrt{5}}{3}$

Step 1: Find the eccentricity $e_1$ and the foci of the hyperbola
The equation of the given hyperbola is:
$\frac{x^2}{16}-\frac{y^2}{9}=1$
Comparing this with the standard equation of a hyperbola $\frac{x^2}{A^2}-\frac{y^2}{B^2}=1$, we get:
$A^2=16\Rightarrow A=4$
$B^2=9\Rightarrow B=3$
The relation between the eccentricity $e_1$ and the semi-axes of the hyperbola is given by:
${e_1}^2=1+\frac{B^2}{A^2}=1+\frac{9}{16}=\frac{25}{16}$
Therefore, the eccentricity of the hyperbola is:
$e_1=\frac{5}{4}$
The coordinates of the foci of this hyperbola are given by $(\pm A{e}_1,0)$:
$(\pm 4·\frac{5}{4},0)=(\pm 5,0)$

Step 2: Determine the parameters of the ellipse
The equation of the ellipse is:
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,a>b$
We are given that the ellipse passes through the foci of the hyperbola, which are $(\pm 5,0)$. Substituting $(5,0)$ into the equation of the ellipse:
$\frac{5^2}{a^2}+\frac{0}{b^2}=1\Rightarrow a^2=25\Rightarrow a=5$
We are also given that:
$e_1{e}_2=1$
Substituting $e_1=\frac{5}{4}$:
$\frac{5}{4}{e}_2=1\Rightarrow e_2=\frac{4}{5}$
For an ellipse with $a>b$, the eccentricity relation is:
${e_2}^2=1-\frac{b^2}{a^2}$
Substituting the values of $e_2$ and $a^2$:
${\left(\frac{4}{5}\right)}^2=1-\frac{b^2}{25}$
$\frac{16}{25}=1-\frac{b^2}{25}$
$\frac{b^2}{25}=1-\frac{16}{25}=\frac{9}{25}\Rightarrow b^2=9$
So, the equation of the ellipse is:
$\frac{x^2}{25}+\frac{y^2}{9}=1$

Step 3: Find the length of the chord parallel to the x-axis passing through (0, 2)
A line parallel to the x-axis and passing through the point $(0,2)$ is given by the equation:
$y=2$
To find the intersection points of this line with the ellipse, we substitute $y=2$ into the equation of the ellipse:
$\frac{x^2}{25}+\frac{2^2}{9}=1$
$\frac{x^2}{25}+\frac{4}{9}=1$
$\frac{x^2}{25}=1-\frac{4}{9}=\frac{5}{9}$
$x^2=\frac{125}{9}$
Taking the square root of both sides gives the x-coordinates of the endpoints of the chord:
$x=\pm \frac{\sqrt{125}}{3}=\pm \frac{5\sqrt{5}}{3}$
The length of the chord is the distance between these two intersection points, which is:
$\text{Length of Chord}=2·\left(\frac{5\sqrt{5}}{3}\right)=\frac{10\sqrt{5}}{3}$