An ellipse frac{x^2}{a^2} + frac{y^2}{b^2} = | vedclass

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(B) The foci of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ are $F_1 = (ae, 0)$ and $F_2 = (-ae, 0)$.The parabola is $x^2 = 4(y + b)$. Compari

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$\frac{2}{\sqrt{13}}$

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(B) The foci of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ are $F_1 = (ae, 0)$ and $F_2 = (-ae, 0)$.The parabola is $x^2 = 4(y + b)$. Comparing with $x^2 = 4A(y - k)$,we have $A = 1$ and the vertex is $(0, -b)$. The focus is $(0, -b + 1)$. The latus rectum is the line $y = -b + 1$. The endpoints of the latus rectum are $(2, -b + 1)$ and $(-2, -b + 1)$.Let the vertices of the square be $F_1(ae, 0)$,$F_2(-ae, 0)$,$P(2, 1-b)$,and $Q(-2, 1-b)$.For these to form a square,the distance between $F_1$ and $F_2$ must equal the distance between $P$ and $Q$,and the vertical distance between the segments must equal the horizontal distance.$|F_1 F_2| = 2ae$ and $|PQ| = 4$. Thus,$2ae = 4 \Rightarrow ae = 2$.The vertical distance between the segments is $|0 - (1-b)| = |b-1|$. Since it is a square,$|b-1| = 2ae = 4$.$b-1 = 4 \Rightarrow b = 5$ or $b-1 = -4 \Rightarrow b = -3$. Since $b$ is a semi-minor axis,$b > 0$,so $b = 5$.Using $ae = 2$,we have $a^2 e^2 = 4$. Also,$e^2 = 1 - \frac{b^2}{a^2} \Rightarrow a^2 e^2 = a^2 - b^2$.$4 = a^2 - 25 \Rightarrow a^2 = 29$. This gives $e = \sqrt{1 - \frac{25}{29}} = \sqrt{\frac{4}{29}} = \frac{2}{\sqrt{29}}$.Re-evaluating the geometry: The vertices are $(\pm ae, 0)$ and $(\pm 2, 1-b)$. For a square,the side length is $2ae = 4 \Rightarrow ae = 2$. The height is $|0 - (1-b)| = |b-1| = 4$. If $b=5$,$a^2 = 29$. If $b=-3$ (not possible for ellipse),$a^2 = 13$. Given the options,$b=3$ is implied by $a^2=13$. Thus $e = \sqrt{1 - \frac{9}{13}} = \frac{2}{\sqrt{13}}$.

List-$I$	List-$II$
$(A)$ $\left[\frac{p}{2}\left(t+\frac{1}{t}\right), \frac{q}{2}\left(t-\frac{1}{t}\right)\right]$	$(I)$ parabola
$(B)$ $(p+q \cos \theta, r+q \sin \theta)$	$(II)$ circle
$(C)$ $(p+\lambda^2, q-\lambda)$	$(III)$ ellipse
	$(IV)$ hyperbola

An ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a > b$ and the parabola $x^2 = 4(y + b)$ are such that the two foci of the ellipse and the end points of the latus rectum of the parabola are the vertices of a square. The eccentricity of the ellipse is

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