Let the hyperbola H : `(x^(2))/(a^(2))-y^(2)=1` and the ellipse `E: 3x^(2)+4y^(2)=12 ` be such that the length of the rectum of H is equal to the length of latus rectum of E . If `e_(H) and e_(E )` are the eccentricities of H and E respectively , then the value of `12(e_(H)^(2)+e_(E )^(2))` is equal to _____.

Answer

To solve the problem, we need to find the value of \( 12(e_H^2 + e_E^2) \), where \( e_H \) and \( e_E \) are the eccentricities of the hyperbola and the ellipse, respectively. We start with the given equations of the hyperbola and the ellipse.

### Step 1: Write the equations in standard form

The hyperbola is given as:
\[ \frac{x^2}{a^2} - y^2 = 1 \]
This is already in standard form.

The ellipse is given as:
\[ 3x^2 + 4y^2 = 12 \]
To convert this into standard form, we divide the entire equation by 12:
\[ \frac{x^2}{4} + \frac{y^2}{3} = 1 \]
Thus, we have:
- For the hyperbola: \( a^2 = a^2 \) and \( b^2 = 1 \)
- For the ellipse: \( a^2 = 4 \) and \( b^2 = 3 \)

### Step 2: Calculate the lengths of the rectum

The length of the rectum of the hyperbola is given by:
\[ L_H = \frac{2b^2}{a} = \frac{2 \cdot 1}{a} = \frac{2}{a} \]

The length of the latus rectum of the ellipse is given by:
\[ L_E = \frac{2b^2}{a} = \frac{2 \cdot 3}{2} = 3 \]

### Step 3: Set the lengths equal

According to the problem, the lengths of the rectum of the hyperbola and the latus rectum of the ellipse are equal:
\[ \frac{2}{a} = 3 \]
Solving for \( a \):
\[ 2 = 3a \implies a = \frac{2}{3} \]

### Step 4: Calculate the eccentricities

Now we calculate the eccentricities of both the hyperbola and the ellipse.

**Eccentricity of the ellipse \( e_E \)**:
\[ e_E = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{3}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]

**Eccentricity of the hyperbola \( e_H \)**:
\[ e_H = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{1}{\left(\frac{2}{3}\right)^2}} = \sqrt{1 + \frac{1}{\frac{4}{9}}} = \sqrt{1 + \frac{9}{4}} = \sqrt{\frac{13}{4}} = \frac{\sqrt{13}}{2} \]

### Step 5: Calculate \( 12(e_H^2 + e_E^2) \)

Now we compute \( e_H^2 \) and \( e_E^2 \):
\[ e_E^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]
\[ e_H^2 = \left(\frac{\sqrt{13}}{2}\right)^2 = \frac{13}{4} \]

Now, we find \( e_H^2 + e_E^2 \):
\[ e_H^2 + e_E^2 = \frac{13}{4} + \frac{1}{4} = \frac{14}{4} = \frac{7}{2} \]

Finally, we calculate \( 12(e_H^2 + e_E^2) \):
\[ 12(e_H^2 + e_E^2) = 12 \cdot \frac{7}{2} = 42 \]

Thus, the final answer is:
\[ \boxed{42} \]