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(B) The extremities of the latus rectum of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ are $(\pm ae, \frac{b^2}{a})$.Since these points lie on

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$x^2 + y^2 = 4$

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(B) The extremities of the latus rectum of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ are $(\pm ae, \frac{b^2}{a})$.Since these points lie on the parabola $x^2 + 2ay - 4 = 0$,we substitute $x = ae$ and $y = \frac{b^2}{a}$ into the equation:$(ae)^2 + 2a(\frac{b^2}{a}) - 4 = 0$$a^2e^2 + 2b^2 - 4 = 0$Using the relation $b^2 = a^2(1 - e^2)$,we substitute $b^2$:$a^2e^2 + 2a^2(1 - e^2) - 4 = 0$$a^2e^2 + 2a^2 - 2a^2e^2 - 4 = 0$$2a^2 - a^2e^2 = 4$$a^2(2 - e^2) = 4 \implies a^2 = \frac{4}{2 - e^2}$Now,find $b^2 = a^2(1 - e^2) = \frac{4(1 - e^2)}{2 - e^2}$.Adding $a^2$ and $b^2$:$a^2 + b^2 = \frac{4}{2 - e^2} + \frac{4(1 - e^2)}{2 - e^2} = \frac{4 + 4 - 4e^2}{2 - e^2} = \frac{4(2 - e^2)}{2 - e^2} = 4$.Thus,the points $(a, b)$ satisfy the equation $x^2 + y^2 = 4$.

If the extremities of the latus rectum having positive ordinate of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ lie on the parabola $x^2 + 2ay - 4 = 0$,then the points $(a, b)$ lie on the curve:

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