1. Fundamental Theorem: Orthogonal Intersection of Conics Implies Confocal Conics

The problem statement begins by telling us that the ellipse and the hyperbola intersect at right angles. This is a crucial piece of information that immediately points to a powerful geometric property:

Theorem: If two conics intersect orthogonally (at right angles), then they are confocal. This means they share the exact same set of foci.

This theorem is the cornerstone of our solution strategy. It allows us to relate the parameters of the ellipse and the hyperbola through their common foci. Our first goal will be to find the foci of the given hyperbola, as these will also be the foci of the ellipse.

2. Analyzing the Hyperbola and Determining its Foci

We are given the equation of the hyperbola: $2x^2 - 2y^2 = 1$

To extract useful information like its eccentricity and foci, we must first convert this equation into the standard form of a hyperbola. The standard form for a hyperbola centered at the origin with its transverse axis along the x-axis is $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$.

Divide the entire given equation by 1 (which is already the constant on the right side) to match the standard form: $\frac{x^2}{1/2} - \frac{y^2}{1/2} = 1$

By comparing this with the standard form, we can identify the parameters for this specific hyperbola. Let's denote them with a subscript 'H' for hyperbola: $A_H^2 = \frac{1}{2} \quad \text{and} \quad B_H^2 = \frac{1}{2}$ From this, we find $A_H = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$. Since $A_H^2 = B_H^2$, this is a special case known as a rectangular hyperbola or equilateral hyperbola.

Next, we need to find its eccentricity, $e_H$. The formula for the eccentricity of a hyperbola is: $e_H = \sqrt{1 + \frac{B_H^2}{A_H^2}}$ We use this formula because it directly relates the eccentricity to the semi-axes of the hyperbola, and $e_H$ must be greater than 1 for a hyperbola. Substituting the values we found: $e_H = \sqrt{1 + \frac{1/2}{1/2}} = \sqrt{1 + 1} = \sqrt{2}$

Now, we can find the foci of this hyperbola. For a hyperbola in the form $\frac{x^2}{A_H^2} - \frac{y^2}{B_H^2} = 1$, the foci are located at $(\pm A_H e_H, 0)$. This is the distance from the center to each focus along the transverse axis. Calculate the focal distance $A_H e_H$: $A_H e_H = \left(\frac{1}{\sqrt{2}}\right)(\sqrt{2}) = 1$ Therefore, the foci of the hyperbola are $(\pm 1, 0)$.

3. Determining the Eccentricity of the Ellipse

The problem statement provides a direct relationship between the eccentricity of the ellipse ($e$) and that of the hyperbola ($e_H$): "the eccentricity of an ellipse is reciprocal to that of the hyperbola" So, we have: $e = \frac{1}{e_H}$ We just calculated $e_H = \sqrt{2}$, so substitute this value: $e = \frac{1}{\sqrt{2}}$ This value makes sense for an ellipse, as its eccentricity must always be between 0 and 1 ($0 < e < 1$).

4. Utilizing the Confocal Property to Find Ellipse Parameters

As established in Step 1, since the ellipse and hyperbola intersect orthogonally, they must be confocal. This means they share the same foci. From Step 2, we found the foci of the hyperbola to be $(\pm 1, 0)$. For an ellipse centered at the origin, with its major axis along the x-axis (which is implied because its foci are on the x-axis, matching the hyperbola's foci), the foci are located at $(\pm ae, 0)$, where $a$ is the semi-major axis and $e$ is the eccentricity of the ellipse.

Therefore, we can equate the focal distances: $ae = 1$ We already found the eccentricity of the ellipse, $e = \frac{1}{\sqrt{2}}$. Substitute this value into the equation: $a \left(\frac{1}{\sqrt{2}}\right) = 1$ Solving for $a$, the semi-major axis of the ellipse: $a = \sqrt{2}$

Now we need to find $b^2$, the square of the semi-minor axis of the ellipse. The relationship between $a, b,$ and $e$ for an ellipse is given by the formula: $e^2 = 1 - \frac{b^2}{a^2}$ This formula is fundamental for ellipses and relates the eccentricity to the lengths of its semi-axes. We use it to find $b^2$ once $a$ and $e$ are known. We know $e = \frac{1}{\sqrt{2}}$, so $e^2 = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}$. We also know $a = \sqrt{2}$, so $a^2 = (\sqrt{2})^2 = 2$. Substitute these values into the eccentricity formula: $\frac{1}{2} = 1 - \frac{b^2}{2}$ Now, we solve for $b^2$: $\frac{b^2}{2} = 1 - \frac{1}{2}$ $\frac{b^2}{2} = \frac{1}{2}$ $b^2 = 1$

5. Calculating the Square of the Length of the Latus Rectum of the Ellipse

The latus rectum of an ellipse is a chord passing through a focus and perpendicular to the major axis. Its length, denoted by $L$, is a key characteristic of the ellipse's shape. For an ellipse with semi-major axis $a$ and semi-minor axis $b$, the length of the latus rectum is given by the formula: $L = \frac{2b^2}{a}$ We have found all the necessary parameters for the ellipse: $b^2 = 1$ and $a = \sqrt{2}$. Substitute these values into the formula: $L = \frac{2(1)}{\sqrt{2}}$ $L = \frac{2}{\sqrt{2}}$ To simplify and rationalize the denominator, we multiply the numerator and denominator by $\sqrt{2}$: $L = \frac{2\sqrt{2}}{2} = \sqrt{2}$ The question asks for the square of the length of the latus rectum. $L^2 = (\sqrt{2})^2 = 2$

Important Tips and Common Pitfalls to Avoid:

• Confocal Property is Key: Always remember and apply the theorem: "If two conics intersect orthogonally, they are confocal." This simplifies many problems involving intersecting conics.
• Eccentricity Formulas: Be extremely careful with the eccentricity formulas for ellipses and hyperbolas.
  • For an ellipse: $e^2 = 1 - \frac{b^2}{a^2}$ (where $a$ is the semi-major axis, $a>b$, and $e<1$).
  • For a hyperbola: $e_H^2 = 1 + \frac{B_H^2}{A_H^2}$ (where $e_H>1$). Note the crucial minus sign for ellipse and plus sign for hyperbola.
• Standard Forms and Parameter Identification: Always convert conic equations to their standard forms to correctly identify $a^2, b^2$ (for ellipse) or $A^2, B^2$ (for hyperbola). For an ellipse $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$, the larger denominator is $a^2$ (semi-major axis squared) and the smaller is $b^2$ (semi-minor axis squared). In this problem, both conics have foci on the x-axis, implying $a$ is associated with the x-term for the ellipse.
• Latus Rectum Formula: Remember the correct formula for the length of the latus rectum of an ellipse is $2b^2/a$. Ensure you use $a$ (semi-major axis) in the denominator.
• Read the Question Carefully: Note that the question asks for the square of the length of the latus rectum, not just the length itself. A common mistake is to stop at $L = \sqrt{2}$.

Conclusion and Key Takeaway:

This problem beautifully demonstrates the power of geometric theorems in coordinate geometry. By recognizing that orthogonally intersecting conics are confocal, we established a common focal length between the ellipse and the hyperbola. Combining this with the given eccentricity relationship, we systematically determined all necessary parameters ($a$ and $b^2$) for the ellipse. Finally, a direct application of the latus rectum formula yielded the required squared length. The key takeaway is to always look for underlying geometric properties, such as the confocal nature, when conics interact in specific ways.