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Step-by-Step Solution
Step 1: Rewrite the Equation of the Hyperbola
The given hyperbola is
$x^2 - y^2 \sec^2 \theta = 10.$
Rewrite it in the standard form
$$
\frac{x^2}{10} - \frac{y^2}{10 \cos^2 \theta} = 1.
$$
From this form, we identify
$a^2 = 10 \quad \text{and} \quad b^2 = 10 \cos^2 \theta.$
Step 2: Find the Eccentricity of the Hyperbola
For the standard hyperbola
$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1,$
its eccentricity $e_H$ is given by
$$
e_H = \sqrt{1 + \frac{b^2}{a^2}}.
$$
Substituting $a^2 = 10$ and $b^2 = 10 \cos^2 \theta,$ we get
$$
e_H
= \sqrt{1 + \frac{10 \cos^2 \theta}{10}}
= \sqrt{1 + \cos^2 \theta}.
$$
Step 3: Rewrite the Equation of the Ellipse
The given ellipse is
$x^2 \sec^2 \theta + y^2 = 5.$
Rewrite it in the form
$$
\frac{x^2}{5 \cos^2 \theta} + \frac{y^2}{5} = 1.
$$
To match the standard ellipse form
$\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$
(noting that the larger denominator corresponds to the major axis),
we see that
$a^2 = 5 \quad \text{and} \quad b^2 = 5 \cos^2 \theta,$
with the major axis along the $y$-axis.
Step 4: Find the Eccentricity of the Ellipse
For an ellipse of the form
$\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$
with $a^2 > b^2,$ the eccentricity $e_E$ is
$$
e_E = \sqrt{1 - \frac{b^2}{a^2}}.
$$
Here, $a^2 = 5$ and $b^2 = 5 \cos^2 \theta,$ giving
$$
e_E
= \sqrt{1 - \frac{5 \cos^2 \theta}{5}}
= \sqrt{1 - \cos^2 \theta}
= \sin \theta.
$$
Step 5: Use the Given Relation Between the Eccentricities
It is given that the eccentricity of the hyperbola is
$\sqrt{5}$
times the eccentricity of the ellipse:
$$
e_H = \sqrt{5}\, e_E.
$$
Substituting
$e_H = \sqrt{1 + \cos^2 \theta}$
and
$e_E = \sqrt{1 - \cos^2 \theta},$
we get
$$
\sqrt{1 + \cos^2 \theta}
= \sqrt{5} \,\sqrt{1 - \cos^2 \theta}.
$$
Squaring both sides,
$$
1 + \cos^2 \theta
= 5 \Bigl(1 - \cos^2 \theta\Bigr).
$$
$$
1 + \cos^2 \theta
= 5 - 5 \cos^2 \theta.
$$
$$
6 \cos^2 \theta
= 4.
$$
$$
\cos^2 \theta
= \frac{2}{3}.
$$
Step 6: Compute the Length of the Latus Rectum of the Ellipse
For an ellipse with major axis along the $y$-axis of the form
$\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1,$
the length of the latus rectum is
$$
L = \frac{2 b^2}{a}.
$$
From our identification:
$$
a = \sqrt{5},
\quad
b^2 = 5 \cos^2 \theta.
$$
Hence,
$$
L
= \frac{2 \bigl(5 \cos^2 \theta\bigr)}{\sqrt{5}}
= \frac{10 \cos^2 \theta}{\sqrt{5}}.
$$
Since $\cos^2 \theta = \tfrac{2}{3},$
$$
L
= \frac{10 \times \frac{2}{3}}{\sqrt{5}}
= \frac{20}{3 \sqrt{5}}
= \frac{20}{3 \sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}}
= \frac{20 \sqrt{5}}{3 \times 5}
= \frac{4 \sqrt{5}}{3}.
$$
Final Answer
The length of the latus rectum of the ellipse is
$$
\frac{4 \sqrt{5}}{3}.
$$
