Let $P(x_{1}, y_{1})$ and $Q(x_{2}, y_{2})$, $y_{1} < 0$, $y_{2} < 0$, be the end points of the latus rectum of the ellipse $x^{2} + 4y^{2} = 4$. The equations of parabolas with latus rectum $PQ$ are

Let P(x₁, y₁) and Q(x₂, y₂), y₁ < 0, y₂ < 0, be the end points of the latus rectum of the ellipse x² + 4y² = 4. The equations of parabolas with latus rectum PQ are

$\text{Given: ellipse } x^2 + 4y^2 = 4$

$P(x_1,y_1),\ Q(x_2,y_2):\ y_1 < 0,\ y_2 < 0$

$PQ = \text{latus rectum of the ellipse}$

$\text{To find: parabolas whose latus rectum is } PQ$

$\text{(select the correct options)}$

$\frac{x^2}{4} + \frac{y^2}{1} = 1$

$\Rightarrow a^2 = 4,\ b^2 = 1 \Rightarrow a = 2,\ b = 1$

$c = \sqrt{a^2-b^2} = \sqrt{4-1} = \sqrt{3}$

$\text{foci: } (\pm\sqrt{3}, 0)$

$\text{latus rectum length} = \frac{2b^2}{a} = \frac{2}{2} = 1$

$\text{at } x = \pm\sqrt{3}:\ y \text{ runs from } -\frac{1}{2} \text{ to } \frac{1}{2}$

$P\left(-\sqrt{3}, -\frac{1}{2}\right),\qquad Q\left(\sqrt{3}, -\frac{1}{2}\right)$

$PQ \text{ is horizontal},\quad y = -\frac{1}{2}$

$M = \left(0, -\frac{1}{2}\right),\qquad PQ = 2\sqrt{3}$

$1.\ \text{midpoint of } PQ = \text{focus of the parabola}$

$2.\ 4a = PQ \Rightarrow a = \frac{\sqrt{3}}{2}$

$3.\ \text{parabolas open up and down}$

$S = \left(0,-\frac{1}{2}\right)$

$4a = PQ = 2\sqrt{3} \Rightarrow a = \frac{\sqrt{3}}{2}$

$k = -\frac{1}{2} - a = -\frac{1+\sqrt{3}}{2}$

$x^2 = 4a\left(y - k\right) = 4\cdot\frac{\sqrt{3}}{2}\left(y + \frac{1+\sqrt{3}}{2}\right)$

$x^2 = 2\sqrt{3}\,y + (3+\sqrt{3})$

$\Rightarrow\ x^2 - 2\sqrt{3}\,y = 3+\sqrt{3}\quad\Rightarrow\ \text{option (2)}$

$k = -\frac{1}{2} + a = \frac{\sqrt{3}-1}{2}$

$x^2 = -4a\left(y - k\right) = -2\sqrt{3}\left(y - \frac{\sqrt{3}-1}{2}\right)$

$x^2 = -2\sqrt{3}\,y + (3-\sqrt{3})$

$\Rightarrow\ x^2 + 2\sqrt{3}\,y = 3-\sqrt{3}\quad\Rightarrow\ \text{option (3)}$

$(1),(4):\ \text{based on } y = +\frac{1}{2}\ (y>0)$

$\text{but the question needs } y < 0\ \Rightarrow\ (1),(4)\ \text{rejected}$

$\boxed{\text{Answer: options (2) and (3)}}$