Let P(x_1, y_1) and Q(x_2, y_2),with y_1

Question

(C) The given ellipse is $\frac{x^2}{4} + \frac{y^2}{1} = 1$. Here $a^2 = 4$ and $b^2 = 1$.The eccentricity $e$ is given by $b^2 = a^2(1 - e^2)$ $\Rig

Vedclass · Accepted Answer

$B, C$

Vedclass · Answer

(C) The given ellipse is $\frac{x^2}{4} + \frac{y^2}{1} = 1$. Here $a^2 = 4$ and $b^2 = 1$.The eccentricity $e$ is given by $b^2 = a^2(1 - e^2)$ $\Rightarrow 1 = 4(1 - e^2)$ $\Rightarrow 1 - e^2 = \frac{1}{4}$ $\Rightarrow e = \frac{\sqrt{3}}{2}$.The coordinates of the foci are $(\pm ae, 0) = (\pm \sqrt{3}, 0)$.Since $P$ and $Q$ are endpoints of the latus rectum with $y Thus,$P = (\sqrt{3}, -\frac{1}{2})$ and $Q = (-\sqrt{3}, -\frac{1}{2})$.The length of the latus rectum $PQ = 2\sqrt{3}$.For a parabola with latus rectum $PQ$,the length of the latus rectum $4a' = 2\sqrt{3} \Rightarrow a' = \frac{\sqrt{3}}{2}$.The midpoint of $PQ$ is $R = (0, -\frac{1}{2})$. The axis of the parabola is the $y$-axis.The vertex of the parabola is $(0, y_v)$,where $y_v = -\frac{1}{2} \pm a' = -\frac{1}{2} \pm \frac{\sqrt{3}}{2}$.Case $1$: Vertex $(0, -\frac{1}{2} - \frac{\sqrt{3}}{2})$. The equation is $x^2 = 4a'(y - y_v)$ $\Rightarrow x^2 = 2\sqrt{3}(y + \frac{1}{2} + \frac{\sqrt{3}}{2})$ $\Rightarrow x^2 = 2\sqrt{3}y + \sqrt{3} + 3$ $\Rightarrow x^2 - 2\sqrt{3}y = 3 + \sqrt{3}$.Case $2$: Vertex $(0, -\frac{1}{2} + \frac{\sqrt{3}}{2})$. The equation is $x^2 = -4a'(y - y_v)$ $\Rightarrow x^2 = -2\sqrt{3}(y + \frac{1}{2} - \frac{\sqrt{3}}{2})$ $\Rightarrow x^2 = -2\sqrt{3}y - \sqrt{3} + 3$ $\Rightarrow x^2 + 2\sqrt{3}y = 3 - \sqrt{3}$.Thus,the equations are $x^2 - 2\sqrt{3}y = 3 + \sqrt{3}$ and $x^2 + 2\sqrt{3}y = 3 - \sqrt{3}$,which correspond to options $B$ and $C$.

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