Question:

Step 1: Identify the coordinates of one end of the latus rectum. For the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$, the latus rectum is a chord perpendicular to the major axis passing through a focus. One end of the latus rectum has coordinates $\left(ae, \dfrac{b^2}{a}\right)$, where $e$ is the eccentricity. Step 2: Write the equation of the normal at a point on the ellipse. The equation of the normal at any point $(x_1, y_1)$ on the ellipse is: $$\frac{a^2 x}{x_1} - \frac{b^2 y}{y_1} = a^2 - b^2$$ Step 3: Substitute the coordinates of the latus rectum endpoint into the normal equation. Substituting $(x_1, y_1) = \left(ae, \dfrac{b^2}{a}\right)$ into the normal equation: $$\frac{a^2 x}{ae} - \frac{b^2 y}{b^2/a} = a^2 - b^2$$ Step 4: Simplify the normal equation. Simplifying the left side: $$\frac{ax}{e} - ay = a^2 - b^2$$ Since $b^2 = a^2(1 - e^2)$, we have $a^2 - b^2 = a^2e^2$. Therefore: $$\frac{ax}{e} - ay = a^2e^2$$ Dividing by $a$: $$\frac{x}{e} - y = ae^2$$ Step 5: Use the condition that the normal passes through an end of the minor axis. The ends of the minor axis are at $(0, b)$ and $(0, -b)$. Substituting the point $(0, -b)$ into the normal equation: $$\frac{0}{e} - (-b) = ae^2$$ $$b = ae^2$$ Step 6: Apply the relationship between $a$, $b$, and $e$. From the ellipse property, $b^2 = a^2(1 - e^2)$, which gives $b = a\sqrt{1-e^2}$. Equating the two expressions for $b$: $$a\sqrt{1-e^2} = ae^2$$ Step 7: Solve for the eccentricity. Dividing both sides by $a$: $$\sqrt{1-e^2} = e^2$$ Squaring both sides: $$1 - e^2 = e^4$$ Rearranging: $$e^4 + e^2 - 1 = 0$$ **Final Answer:** The relationship between the eccentricity and the given condition is $e^4 + e^2 - 1 = 0$. This matches **Option 4**.