Question #58fdd

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Question #58fdd

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Answer 1 · 2017-04-13T19:06:24

$(\pm \frac{\sqrt{3}}{2}, 0) .$ $(+-sqrt3/2,0).$

Explanation:

$x^{2} = 1 - 4 y^{2} \Rightarrow x^{2} + 4 y^{2} = 1, or, x^{2} + \frac{y^{2}}{\frac{1}{4}} = 1$ $x^2=1-4y^2 rArr x^2+4y^2=1, or, x^2+y^2/(1/4)=1$

Comparing with the Standard Eqn. of Ellipse, i.e.,

$\frac{x^{2}}{a^{2}} + \frac{t^{2}}{b^{2}} = 1,$ $x^2/a^2+t^2/b^2=1,$ we have, $a^{2} = 1, b^{2} = \frac{1}{4}, s o, a > b .$ $a^2=1, b^2=1/4, so, a >b.$

Knowing that the Eccentricity $e$ $e$ is given by,

$b^{2} = a^{2} (1 - e^{2}),$ $b^2=a^2(1-e^2),$ we get, $\frac{1}{4} = 1 (1 - e^{2})$ $1/4=1(1-e^2)$

$:. e=sqrt3/2.$

Therefore, the Focii are, $(+-ae,0)=(+-sqrt3/2,0).$

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Question #58fdd

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