The value of a for the ellipse $$\dfrac{x^{2}}{a^{2}}+ \dfrac{y^{2}}{b^{2}}=1 $$(a>b) , if the extremities of the latus rectum of the ellipse having positive ordinate lies on the parabola $$x^{2}= -2 (y-2)$$ is
$$(\pm ae, \dfrac{b^{2}}{a})$$ are extremities of the latus - rectum having positive ordinates .
$$a^{2}e^{2}= -2(\dfrac{b^{2}}{a}- 2)$$
But $$b^{2}= a^{2}(1-e^{2})$$
From (1) and (2) , we get $$a^{2}e^{2}- 2ae^{2} + 2a -4=0 $$
$$ae^{2}(a-2) + 2(a-2)=0$$
$$(ae^{2}+ 2)(a-2)=0$$
Hence a=2.
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