foci ((x+3)^2}{25}-\frac{(y-4)^2)/9 =1

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

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foci (x+3)²25 −(y−4)²9 =1

Solution

(−3+√34, 4), (−3−√34, 4)

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Hyperbola Foci

For a right-left facing hyperbola, the Foci (focus points) are defined as (h+c, k), (h−c, k),
where c=√a²+b² is the distance from the center (h, k) to a focus

(h+c, k), (h−c, k)

Calculate Hyperbola properties

(x+3)²25 −(y−4)²9 =1: Right-left Hyperbola with (h, k)=(−3, 4), a=5, b=3

(−3+c, 4), (−3−c, 4)

Compute c:

c=√5²+3²: √34

(−3+√34, 4), (−3−√34, 4)

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

foci ((x+3)^2}{25}-\frac{(y-4)^2)/9 =1

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