This problem requires a solid understanding of the geometric properties of an ellipse, specifically the coordinates of its key points and the equation of its normal. We will systematically apply these concepts to derive the relationship between the semi-axes and then express it in terms of the eccentricity.

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1. Essential Concepts and Formulas for an Ellipse

Let's consider a standard ellipse centered at the origin $(0,0)$ with its major axis along the x-axis.

• Standard Equation of an Ellipse: The equation is given by: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a$ is the length of the semi-major axis (half the major axis) and $b$ is the length of the semi-minor axis (half the minor axis). For an ellipse with its major axis along the x-axis, we have the condition $a > b$.

• Relationship between Semi-axes and Eccentricity ($e$): The eccentricity $e$ is a fundamental characteristic of an ellipse, quantifying its "ovalness." It always satisfies $0 < e < 1$. The relationship between $a$, $b$, and $e$ is given by: $b^2 = a^2(1-e^2)$ This formula is crucial for converting expressions involving $a$ and $b$ into expressions involving $e$.

• Coordinates of the Ends of the Latus Rectum: The latus rectum is a chord passing through a focus and perpendicular to the major axis. The foci of the ellipse are located at $(\pm ae, 0)$. The four ends of the latus rectum are: $\left(ae, \pm \frac{b^2}{a}\right) \quad \text{and} \quad \left(-ae, \pm \frac{b^2}{a}\right)$ For our problem, we will choose one of these points, say $P(x_1, y_1) = \left(ae, \frac{b^2}{a}\right)$, which lies in the first quadrant. Due to the symmetry of the ellipse, the final result for eccentricity $e$ will be independent of which specific end of the latus rectum we choose.

• Coordinates of the Extremities of the Minor Axis: The minor axis lies along the y-axis, and its endpoints are: $(0, \pm b)$

• Equation of the Normal to an Ellipse at a point $(x_1, y_1)$: The equation of the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at a point $P(x_1, y_1)$ on the ellipse is given by: $\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2$ Why this formula? The slope of the tangent at $(x_1, y_1)$ is obtained by implicit differentiation: $\frac{2x}{a^2} + \frac{2y}{b^2}\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{b^2x_1}{a^2y_1}$. Since the normal is perpendicular to the tangent, its slope is $m_N = \frac{-1}{dy/dx} = \frac{a^2y_1}{b^2x_1}$. Using the point-slope form $y - y_1 = m_N(x - x_1)$ and rearranging terms leads directly to the standard normal equation provided above.

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2. Step-by-Step Solution

Step 2.1: Identify the Specific Points Involved The problem states that the normal at an end of a latus rectum passes through an extremity of the minor axis.

• Point on the Ellipse ($P(x_1, y_1)$): We select an end of the latus rectum. For simplicity and to work with positive coordinates initially, we choose the point in the first quadrant: $P \left(x_1, y_1\right) = \left(ae, \frac{b^2}{a}\right)$ This point lies on the ellipse by definition.

• Point through which the Normal Passes ($Q(x_0, y_0)$): The normal passes through an extremity of the minor axis. The minor axis extremities are $(0, b)$ and $(0, -b)$. Let's consider the geometry: if the point $P$ is in the first quadrant ($y_1 > 0$), the normal typically slopes "inward" towards the major axis. For it to pass through a minor axis extremity, it's geometrically more plausible for it to pass through the one on the opposite side of the major axis, i.e., $(0, -b)$. $Q(x_0, y_0) = (0, -b)$ Self-check: If we were to choose $Q(0, b)$, the equation would lead to $ab = -(a^2-b^2)$, which after squaring would yield the same result for $e$. However, selecting $(0, -b)$ aligns better with the typical visual representation of a normal from the first quadrant.

Step 2.2: Write the Equation of the Normal at Point $P(x_1, y_1)$ Substitute the coordinates of $P(x_1, y_1) = \left(ae, \frac{b^2}{a}\right)$ into the general normal equation $\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2$: $\frac{a^2x}{ae} - \frac{b^2y}{b^2/a} = a^2 - b^2$ Now, simplify the terms: $\frac{ax}{e} - ay = a^2 - b^2$ This is the specific equation of the normal line for our chosen point $P$.

Step 2.3: Apply the Condition that the Normal Passes Through $Q(0, -b)$ Since the normal line must pass through $Q(0, -b)$, the coordinates of $Q$ must satisfy the normal's equation. Substitute $x=0$ and $y=-b$ into the equation from Step 2.2: $\frac{a(0)}{e} - a(-b) = a^2 - b^2$ $0 + ab = a^2 - b^2$ $ab = a^2 - b^2$ This equation establishes the fundamental relationship between $a$ and $b$ based on the problem's geometric condition.

Step 2.4: Express the Equation in Terms of Eccentricity ($e$) Our goal is to find a condition for the eccentricity $e$. We need to eliminate $a$ and $b$ from the equation $ab = a^2 - b^2$ using the relationship $b^2 = a^2(1-e^2)$.

• Normalize the equation: To simplify substitution, it's often helpful to work with ratios. Divide the entire equation $ab = a^2 - b^2$ by $a^2$ (we can do this because $a \neq 0$ for a non-degenerate ellipse): $\frac{ab}{a^2} = \frac{a^2}{a^2} - \frac{b^2}{a^2}$ $\frac{b}{a} = 1 - \frac{b^2}{a^2}$

• Substitute using eccentricity: From the relationship $b^2 = a^2(1-e^2)$, we can derive: $\frac{b^2}{a^2} = 1-e^2$ Taking the square root of both sides (and noting that $b/a > 0$ and $1-e^2 > 0$ since $0 < e < 1$): $\frac{b}{a} = \sqrt{1-e^2}$ Now, substitute these expressions for $b/a$ and $b^2/a^2$ into the normalized equation: $\sqrt{1-e^2} = 1 - (1-e^2)$ $\sqrt{1-e^2} = 1 - 1 + e^2$ $\sqrt{1-e^2} = e^2$ This simplified equation is solely in terms of $e$.

Step 2.5: Solve for $e$ To eliminate the square root, we square both sides of the equation $\sqrt{1-e^2} = e^2$: $\left(\sqrt{1-e^2}\right)^2 = (e^2)^2$ $1-e^2 = e^4$ Finally, rearrange the terms to form a polynomial equation in $e$: $e^4 + e^2 - 1 = 0$ This is the required condition that the eccentricity $e$ must satisfy.

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3. Important Considerations and Common Pitfalls

• Correct Normal Equation: Double-check the formula for the normal. A common error is to confuse it with the tangent equation or to misplace terms.
• Geometric Intuition for Point Selection: While choosing any end of the latus rectum and any extremity of the minor axis will ultimately lead to the same algebraic condition for $e$ (due to squaring, which handles sign differences), having a consistent geometric visualization (e.g., normal from $P(ae, b^2/a)$ passing through $Q(0, -b)$) can help in avoiding intermediate sign errors and provides confidence in the setup.
• Algebraic Precision: Be meticulous with algebraic manipulations, especially when substituting $b^2 = a^2(1-e^2)$. It's often safer to first get ratios like $b^2/a^2$ and $b/a$ before substitution.
• Squaring and Extraneous Solutions: When squaring both sides of an equation (as in $\sqrt{1-e^2} = e^2$), it's crucial to consider if extraneous solutions might be introduced. In this specific case, for $e \in (0,1)$, $e^2$ is positive, and $\sqrt{1-e^2}$ is defined as the positive square root. Therefore, both sides of $\sqrt{1-e^2} = e^2$ are positive, and squaring does not introduce extraneous solutions for $e^2$ that are valid within the domain of $e$. For example, $e^2 = \frac{-1 \pm \sqrt{1-4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}$. Since $e^2$ must be positive, $e^2 = \frac{-1 + \sqrt{5}}{2}$ is the only valid solution.

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4. Summary and Key Takeaway

This problem is an excellent exercise in applying the fundamental properties of an ellipse to solve a geometric condition. The solution hinges on these key steps:

1. Accurate Identification: Correctly identifying the coordinates of the relevant points (end of latus rectum and extremity of minor axis).
2. Formula Application: Using the correct equation for the normal to the ellipse.
3. Condition Translation: Substituting the coordinates of the second point into the normal equation to translate the geometric condition into an algebraic relationship between $a$ and $b$.
4. Eccentricity Conversion: Employing the defining relationship $b^2 = a^2(1-e^2)$ to convert the algebraic relationship into an equation solely involving the eccentricity $e$.

The final condition for the eccentricity $e$ is $\boxed{\text{e 4 + e 2 – 1 = 0}}$.