This problem is a classic example that tests your comprehensive understanding of conic sections, specifically ellipses and hyperbolas. We are given an ellipse with certain properties (eccentricity and latus rectum length) and asked to find the eccentricity of a related hyperbola. The strategy involves using the given information about the ellipse to determine its fundamental parameters ($a^2$ and $b^2$), and then applying these parameters to calculate the eccentricity of the specified hyperbola.

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1. Key Concepts and Formulas

Let's first define the standard forms and essential properties of the conic sections relevant to this problem.

For an Ellipse: The standard equation of an ellipse centered at the origin is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

• In this problem, we are explicitly given $a > b$. This means the major axis lies along the x-axis, $a$ is the semi-major axis, and $b$ is the semi-minor axis.
• Eccentricity ($e$): A measure of how "flat" or "stretched" an ellipse is. For an ellipse with $a > b$, its eccentricity is given by: $e = \sqrt{1 - \frac{b^2}{a^2}}$ An alternative and very useful form derived from this is $b^2 = a^2(1-e^2)$. Remember that for an ellipse, $0 < e < 1$.
• Length of Latus Rectum ($L$): This is the length of the chord passing through a focus and perpendicular to the major axis. For an ellipse with $a > b$, the formula is: $L = \frac{2b^2}{a}$

For a Hyperbola: The standard equation of a hyperbola centered at the origin with its transverse axis along the x-axis is $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$.

• Here, $A$ is the length of the semi-transverse axis, and $B$ is the length of the semi-conjugate axis.
• Eccentricity ($e'$): A measure of the "openness" of the hyperbola's branches. The formula is: $e' = \sqrt{1 + \frac{B^2}{A^2}}$ The square of the eccentricity, $(e')^2$, which is often what is directly calculated, is $(e')^2 = 1 + \frac{B^2}{A^2}$. Note that for a hyperbola, $e' > 1$.
  • Important Note for this problem: The hyperbola given is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. This means that the $A^2$ and $B^2$ of the hyperbola are directly equivalent to the $a^2$ and $b^2$ values that we will determine from the ellipse. So, for this specific hyperbola, $A^2 = a^2$ and $B^2 = b^2$.

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2. Step-by-Step Solution

We will break down the problem into two main parts: first, finding the parameters of the ellipse, and then using those to find the eccentricity of the hyperbola.

Part 1: Determining Parameters of the Ellipse

We are given the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with the condition $a > b$.

Step 1: Utilize the given eccentricity of the ellipse. The eccentricity of the ellipse is given as $e = \frac{1}{\sqrt{2}}$. The formula for the eccentricity of an ellipse with $a > b$ is: $e = \sqrt{1 - \frac{b^2}{a^2}}$ Substitute the given value of $e$ into this formula: $\frac{1}{\sqrt{2}} = \sqrt{1 - \frac{b^2}{a^2}}$ Why this step? The eccentricity provides a fundamental algebraic relationship between the semi-major axis ($a$) and the semi-minor axis ($b$) of the ellipse. Establishing this relationship is the first crucial step towards finding the specific values of $a$ and $b$.

Step 2: Establish a relationship between $a^2$ and $b^2$. To simplify the equation from Step 1 and eliminate the square root, we square both sides: $\left(\frac{1}{\sqrt{2}}\right)^2 = \left(\sqrt{1 - \frac{b^2}{a^2}}\right)^2$ $\frac{1}{2} = 1 - \frac{b^2}{a^2}$ Now, rearrange the terms to isolate the ratio $\frac{b^2}{a^2}$: $\frac{b^2}{a^2} = 1 - \frac{1}{2}$ $\frac{b^2}{a^2} = \frac{1}{2}$ This gives us a direct relationship between $a^2$ and $b^2$: $a^2 = 2b^2 \quad \text{(Equation 1)}$ Why this step? By simplifying the eccentricity equation, we obtain a clear algebraic connection between $a^2$ and $b^2$. This allows us to express one variable in terms of the other, which will be essential when we use the second piece of information (latus rectum length) to solve for their unique values.

Step 3: Incorporate the given length of the latus rectum. The length of the latus rectum ($L$) for an ellipse with $a > b$ is given by: $L = \frac{2b^2}{a}$ We are given that $L = \sqrt{14}$. Substitute this value into the formula: $\sqrt{14} = \frac{2b^2}{a}$ Why this step? The length of the latus rectum is another geometric property that depends on $a$ and $b$. By introducing this second equation, we now have a system of two independent equations with two unknowns ($a$ and $b$), which can be solved uniquely.

Step 4: Solve for $a^2$ and $b^2$. From Equation 1, we have $a^2 = 2b^2$. Since $a$ and $b$ represent lengths, they must be positive, so we can write $a = \sqrt{2b^2} = b\sqrt{2}$. Substitute this expression for $a$ into the latus rectum equation from Step 3: $\sqrt{14} = \frac{2b^2}{b\sqrt{2}}$ Simplify the right side: $\sqrt{14} = \frac{2b}{\sqrt{2}}$ Since $2 = (\sqrt{2})^2$, we can write $\frac{2}{\sqrt{2}} = \sqrt{2}$: $\sqrt{14} = b\sqrt{2}$ Now, solve for $b$: $b = \frac{\sqrt{14}}{\sqrt{2}} = \sqrt{\frac{14}{2}} = \sqrt{7}$ Squaring $b$, we get the value of $b^2$: $b^2 = 7$ Finally, use Equation 1 ($a^2 = 2b^2$) to find $a^2$: $a^2 = 2(7)$ $a^2 = 14$ Why this step? We have now successfully determined the specific numerical values of $a^2$ and $b^2$ for the given ellipse. These are the fundamental parameters that will be directly used to define the hyperbola in the next part of the problem.

Part 2: Calculating the Eccentricity of the Hyperbola

We need to find the square of the eccentricity of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

Step 5: Identify the hyperbola's parameters and its eccentricity formula. The given hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Comparing this to the standard form of a hyperbola with transverse axis along the x-axis, $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$, we can directly identify its parameters: $A^2 = a^2 \quad \text{and} \quad B^2 = b^2$ The formula for the square of the eccentricity $(e')^2$ of such a hyperbola is: $(e')^2 = 1 + \frac{B^2}{A^2}$ Substituting $A^2 = a^2$ and $B^2 = b^2$ (from the context of this problem): $(e')^2 = 1 + \frac{b^2}{a^2}$ Why this step? It is crucial to correctly identify the correspondence between the parameters ($a^2, b^2$) derived from the ellipse and the parameters ($A^2, B^2$) of the hyperbola, as specified in the problem statement. This ensures that the correct values are substituted into the appropriate hyperbola eccentricity formula.

Step 6: Substitute the determined values of $a^2$ and $b^2$ and calculate $(e')^2$. From Part 1, we found $a^2 = 14$ and $b^2 = 7$. Substitute these values into the eccentricity formula for the hyperbola: $(e')^2 = 1 + \frac{7}{14}$ Simplify the fraction: $(e')^2 = 1 + \frac{1}{2}$ Perform the addition: $(e')^2 = \frac{2}{2} + \frac{1}{2}$ $(e')^2 = \frac{3}{2}$ Why this step? This is the final calculation that directly answers the question posed. By substituting the previously determined values, we obtain the required square of the eccentricity for the hyperbola.

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3. Tips and Common Mistakes

• Distinguish Ellipse vs. Hyperbola Formulas: This is the most critical point. A very common error is mixing up the eccentricity formulas due to the similar appearance. Remember the key difference:
  • Ellipse: $e = \sqrt{1 - \frac{\text{minor axis}^2}{\text{major axis}^2}}$ (subtraction, $0 < e < 1$)
  • Hyperbola: $e' = \sqrt{1 + \frac{\text{conjugate axis}^2}{\text{transverse axis}^2}}$ (addition, $e' > 1$)
• Definition of $a, b, A, B$: For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, $a$ always represents the semi-major axis and $b$ the semi-minor axis, meaning $a > b$ (if major axis is along x-axis) or $b > a$ (if major axis is along y-axis). However, for a hyperbola $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$, $A$ is the semi-transverse axis and $B$ is the semi-conjugate axis; there is no general rule that $A > B$ or $B > A$. In this specific problem, the hyperbola uses the same symbols ($a^2, b^2$) as the ellipse, which simplifies the parameter mapping but can also be a source of confusion if not handled carefully. Always ensure you are using the correct parameters for the specific conic section in question.
• Algebraic Precision: Be meticulous with squaring, taking square roots, and fraction arithmetic. A small calculation error, such as simplifying $\frac{2b^2}{b\sqrt{2}}$ incorrectly, can lead to an incorrect final answer.

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4. Summary and Key Takeaway

We successfully determined the square of the eccentricity of the hyperbola by systematically leveraging the given properties of the ellipse. The process involved:

1. Using the ellipse's eccentricity to find a fundamental algebraic relationship between $a^2$ and $b^2$.
2. Using the ellipse's latus rectum length, along with the previously found relationship, to solve for the specific numerical values of $a^2$ and $b^2$.
3. Applying these determined $a^2$ and $b^2$ values to the hyperbola's eccentricity formula, ensuring correct mapping of parameters.

The key takeaway is that a strong understanding of the definitions and formulas for conic sections, combined with a systematic, step-by-step application, is crucial for solving such problems. Always pay close attention to the signs in the