If $H:\frac{x^2}{16}-<!--\; -\; -->\frac{y^2}{9}=1$ and $E:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b).$ Ellipse passes through the foci of the hyperbola and e₁. e₂ = 1 (where e₁, e₂ are the eccentricities of hyperbola and ellipse, respectively). The length of the chord of ellipse passing through (0, 2) is equal to

Correct Answer: 10√5/3 (which corresponds to the option 10√5/3)

Here is the step-by-step explanation and derivation to solve the problem:

Step 1: Find the eccentricity and foci of the given hyperbola
The equation of the hyperbola is given as:

$H:\frac{x^2}{16}-\frac{y^2}{9}=1$

Comparing this with the standard hyperbola equation $\frac{x^2}{A^2}-\frac{y^2}{B^2}=1$, we find:
$A^2=16$ and $B^2=9$.
The eccentricity e₁ of the hyperbola is given by:

$e_1^2=1+\frac{B^2}{A^2}=1+\frac{9}{16}=\frac{25}{16}$

Taking the square root, we get:
$e_1=\frac{5}{4}$
The coordinates of the foci of the hyperbola are $(\pm A{e}_1,0)$:

$\text{Foci}=(\pm 4\times \frac{5}{4},0)=(\pm 5,0)$

Step 2: Determine the equation of the ellipse
We are given that the ellipse passes through the foci of the hyperbola. Looking at the provided image illustration, the ellipse is centered at the origin with its major axis along the x-axis (since a > b).
Since the ellipse passes through the foci $(\pm 5,0)$, these points must be the horizontal vertices of the ellipse:
$a=5$
We are also given that the relation between the eccentricities is:

$e_1\cdot e_2=1$

Substituting $e_1=\frac{5}{4}$, we obtain the eccentricity of the ellipse e₂:
$e_2=\frac{4}{5}$
For the ellipse, we use the standard eccentricity relation:

$1-\frac{b^2}{a^2}=e_2^2$

Substituting the values of a and e₂:

$1-\frac{b^2}{25}={\left(\frac{4}{5}\right)}^2=\frac{16}{25}$

$\frac{b^2}{25}=1-\frac{16}{25}=\frac{9}{25}$

This gives:
$b^2=9$
Thus, the equation of the ellipse is:

$E:\frac{x^2}{25}+\frac{y^2}{9}=1$

Step 3: Calculate the length of the chord passing through (0, 2)
The question asks for the length of the chord of the ellipse passing through $(0,2)$. As shown in the step-by-step solution in the image, this is the horizontal chord along the line $y=2$.
Substituting $y=2$ into the ellipse equation:

$\frac{x^2}{25}+\frac{2^2}{9}=1$

$\frac{x^2}{25}=1-\frac{4}{9}=\frac{5}{9}$

Multiplying by 25:
$x^2=\frac{125}{9}$
Taking the square root:

$x=\pm \frac{5\sqrt{5}}{3}$

The coordinates of the endpoints of the chord are:
$x_1=-\frac{5\sqrt{5}}{3}$ and $x_2=\frac{5\sqrt{5}}{3}$.
Therefore, the total length l of the chord is:

$l=x_2-x_1=2\times \frac{5\sqrt{5}}{3}=\frac{10\sqrt{5}}{3}$