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foci x^2+y^2=1

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

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ddx

≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
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foci x²+y²=1

Solution

(0, 0), (0, 0)

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Ellipse Foci

For an ellipse with major axis parallel to the y-axis, the Foci (focus points) are defined as (h, k+c), (h, k−c),
where c=√b²−a² is the distance from the center (h, k) to a focus

(h, k+c), (h, k−c)

Calculate ellipse properties

x²+y²=1: Ellipse with center (h, k)=(0, 0), b=1, a=1

(0, 0+c), (0, 0−c)

Compute c:

c=√1²−1²: 0

(0, 0+0), (0, 0−0)

Refine

(0, 0), (0, 0)

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

foci x^2+y^2=1

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