Step 1: Analyze the given ellipse

The equation of the ellipse is given as ${x^2} + 4{y^2} = 4$. To convert this to the standard form ${\frac{x^2}{a^2}} + {\frac{y^2}{b^2}} = 1$, we divide the entire equation by 4: {\frac{x^2}{4}} + {\frac{4y^2}{4}} = {\frac{4}{4}} $$$$ {\frac{x^2}{4}} + {\frac{y^2}{1}} = 1 From the standard equation, we can identify: $a^2 = 4 \implies a = 2$ $b^2 = 1 \implies b = 1$ Since $a > b$, the major axis of the ellipse is along the x-axis.

Step 2: Find the foci and endpoints of the latus recta of the ellipse

The eccentricity $e$ is given by the formula $b^2 = a^2(1 - e^2)$. 1 = 4(1 - e^2) $$$$ \frac{1}{4} = 1 - e^2 $$$$ e^2 = 1 - \frac{1}{4} = \frac{3}{4} $$$$ e = \frac{\sqrt{3}}{2} The foci of the ellipse are at $(\pm ae, 0)$. $ae = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}$ So, the foci are at $S_1(\sqrt{3}, 0)$ and $S_2(-\sqrt{3}, 0)$.

The endpoints of the latus recta are given by $(\pm ae, \pm \frac{b^2}{a})$. $\frac{b^2}{a} = \frac{1}{2}$ The four endpoints of the latus recta are $(\sqrt{3}, \frac{1}{2})$, $(\sqrt{3}, -\frac{1}{2})$, $(-\sqrt{3}, \frac{1}{2})$, and $(-\sqrt{3}, -\frac{1}{2})$.

Step 3: Identify the points P and Q

The question states that $P(x_1, y_1)$ and $Q(x_2, y_2)$ are the endpoints of the latus rectum, with the conditions $y_1 < 0$ and $y_2 < 0$. A single latus rectum of the ellipse is a vertical chord, so its endpoints have the same x-coordinate and y-coordinates with opposite signs (e.g., $(\sqrt{3}, 1/2)$ and $(\sqrt{3}, -1/2)$). This contradicts the condition that both $y_1$ and $y_2$ are negative. Therefore, the problem must be interpreted as P and Q being the two endpoints of the latus recta that lie in the lower half-plane (where y is negative). So, the coordinates of P and Q are: $P(\sqrt{3}, -1/2) \quad \text{and} \quad Q(-\sqrt{3}, -1/2)$

Step 4: Define the properties of the parabolas

The segment PQ is the latus rectum for the two required parabolas.

• Length of the latus rectum (4A): The distance between P and Q. $4A = |x_1 - x_2| = |\sqrt{3} - (-\sqrt{3})| = 2\sqrt{3}$ $A = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}$
• Axis of the parabola: The axis is perpendicular to the latus rectum and passes through its midpoint. PQ is a horizontal segment on the line $y = -1/2$. Therefore, the axis of the parabolas must be vertical. The midpoint of PQ is $(\frac{\sqrt{3}-\sqrt{3}}{2}, \frac{-1/2-1/2}{2}) = (0, -1/2)$. So, the axis of the parabolas is the y-axis, i.e., the line $x=0$.
• Focus of the parabola (S): The focus is the intersection of the axis and the line containing the latus rectum. Thus, the focus S is at $(0, -1/2)$.

Step 5: Derive the equations of the two possible parabolas

The vertex of a parabola is at a distance $A$ from the focus along its axis. Since the axis is $x=0$, the vertex $(h, k)$ will be $(0, -1/2 \pm A)$.

Case 1: Parabola opens upwards The vertex V is below the focus S. $V_1 = (0, -1/2 - A) = \left(0, -\frac{1}{2} - \frac{\sqrt{3}}{2}\right) = \left(0, -\frac{1+\sqrt{3}}{2}\right)$ The equation of an upward-opening parabola is $(x-h)^2 = 4A(y-k)$. (x-0)^2 = 2\sqrt{3} \left( y - \left(-\frac{1+\sqrt{3}}{2}\right) \right) $$$$ x^2 = 2\sqrt{3} \left( y + \frac{1+\sqrt{3}}{2} \right) $$$$ x^2 = 2\sqrt{3}y + \sqrt{3}(1+\sqrt{3}) $$$$ x^2 = 2\sqrt{3}y + \sqrt{3} + 3 $$$$ x^2 - 2\sqrt{3}y = 3 + \sqrt{3} This corresponds to option B.

Case 2: Parabola opens downwards The vertex V is above the focus S. $V_2 = (0, -1/2 + A) = \left(0, -\frac{1}{2} + \frac{\sqrt{3}}{2}\right) = \left(0, \frac{\sqrt{3}-1}{2}\right)$ The equation of a downward-opening parabola is $(x-h)^2 = -4A(y-k)$. (x-0)^2 = -2\sqrt{3} \left( y - \left(\frac{\sqrt{3}-1}{2}\right) \right) $$$$ x^2 = -2\sqrt{3}y + \sqrt{3}(\sqrt{3}-1) $$$$ x^2 = -2\sqrt{3}y + 3 - \sqrt{3} $$$$ x^2 + 2\sqrt{3}y = 3 - \sqrt{3} This corresponds to option C.

Conclusion

The equations of the two possible parabolas are $x^2 - 2\sqrt{3}y = 3 + \sqrt{3}$ and $x^2 + 2\sqrt{3}y = 3 - \sqrt{3}$. Thus, options B and C are correct.