If `x^2-y^2sec^2theta=10` be a hyperbola and `x^2sec^2theta+y^2=5` be an ellipse such that the eccentricity of hyperbola=`sqrt(5)` eccentricity of ellipse then find the length of latus rectum of ellipse

Options

A. `(4sqrt(5))/3`
B. `4/(3sqrt(5))`
C. `(20sqrt(5))/3`
D. `sqrt(30)`

Answer

To solve the problem, we need to analyze the given equations of the hyperbola and the ellipse, and then find the length of the latus rectum of the ellipse.

### Step 1: Analyze the Hyperbola
The equation of the hyperbola is given as:
\[ x^2 - y^2 \sec^2 \theta = 10 \]
We can rewrite this in standard form:
\[ \frac{x^2}{10} - \frac{y^2}{10 \cos^2 \theta} = 1 \]
Here, we identify \( a^2 = 10 \) and \( b^2 = 10 \cos^2 \theta \).

### Step 2: Calculate the Eccentricity of the Hyperbola
The eccentricity \( e_h \) of the hyperbola is given by:
\[ e_h = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{10 \cos^2 \theta}{10}} = \sqrt{1 + \cos^2 \theta} \]
According to the problem, \( e_h = \sqrt{5} \). Thus, we have:
\[ \sqrt{1 + \cos^2 \theta} = \sqrt{5} \]
Squaring both sides gives:
\[ 1 + \cos^2 \theta = 5 \implies \cos^2 \theta = 4 \implies \cos \theta = \pm 2 \]
However, since \( \cos \theta \) must be between -1 and 1, we need to ensure that our calculations are consistent with the conditions of the problem.

### Step 3: Analyze the Ellipse
The equation of the ellipse is given as:
\[ x^2 \sec^2 \theta + y^2 = 5 \]
Rearranging gives:
\[ \frac{x^2}{5 \cos^2 \theta} + \frac{y^2}{5} = 1 \]
Here, we identify \( a^2 = 5 \) and \( b^2 = 5 \cos^2 \theta \).

### Step 4: Calculate the Eccentricity of the Ellipse
The eccentricity \( e_e \) of the ellipse is given by:
\[ e_e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{5}{5 \cos^2 \theta}} = \sqrt{1 - \frac{1}{\cos^2 \theta}} = \sqrt{\frac{\sin^2 \theta}{\cos^2 \theta}} = \tan \theta \]
We are given that \( e_h = e_e \), which implies:
\[ \sqrt{1 + \cos^2 \theta} = \tan \theta \]

### Step 5: Solve for \( \theta \)
From the previous equations, equate:
\[ \sqrt{1 + \cos^2 \theta} = \tan \theta \]
Squaring both sides gives:
\[ 1 + \cos^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} \]
Substituting \( \sin^2 \theta = 1 - \cos^2 \theta \):
\[ 1 + \cos^2 \theta = \frac{1 - \cos^2 \theta}{\cos^2 \theta} \]
Multiplying through by \( \cos^2 \theta \):
\[ \cos^2 \theta + \cos^4 \theta = 1 - \cos^2 \theta \]
Rearranging gives:
\[ \cos^4 \theta + 2\cos^2 \theta - 1 = 0 \]
Let \( u = \cos^2 \theta \):
\[ u^2 + 2u - 1 = 0 \]
Using the quadratic formula:
\[ u = \frac{-2 \pm \sqrt{4 + 4}}{2} = -1 \pm \sqrt{2} \]
Since \( u \) must be non-negative, we take:
\[ u = -1 + \sqrt{2} \implies \cos^2 \theta = -1 + \sqrt{2} \]

### Step 6: Length of the Latus Rectum of the Ellipse
The length of the latus rectum \( L \) of the ellipse is given by:
\[ L = \frac{2b^2}{a} \]
Substituting \( a = \sqrt{5} \) and \( b = \sqrt{5 \cos^2 \theta} \):
\[ L = \frac{2(5 \cos^2 \theta)}{\sqrt{5}} = \frac{10 \cos^2 \theta}{\sqrt{5}} \]
Substituting \( \cos^2 \theta = -1 + \sqrt{2} \):
\[ L = \frac{10(-1 + \sqrt{2})}{\sqrt{5}} = \frac{10\sqrt{2} - 10}{\sqrt{5}} \]

### Final Answer
The length of the latus rectum of the ellipse is:
\[ L = \frac{10(\sqrt{2} - 1)}{\sqrt{5}} \]