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

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simplify solve for inverse tangent line

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axis (x−4)²+(y+2)²=25

Solution

semi−major axis b=5, semi−minor axis a=5

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(x−4)²+(y+2)²=25

Ellipse standard equation

(x−h)²a² +(y−k)²b² =1 is the ellipse standard equation
with center (h, k) and a, b are the semi-major and semi-minor axes

Rewrite (x−4)²+(y+2)²=25 in the form of the standard ellipse equation

(x−4)²5² +(y−(−2))²5² =1

Therefore ellipse properties are:

(h, k)=(4, −2), a=5, b=5

b>a therefore b is semi-major axis and a is semi-minor axis

Ellipse with center (h, k)=(4, −2), b=5, a=5

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

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1	−4
4	−7