trục 25x^2+4y^2+100x-40y=400

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axis 25x²+4y²+100x−40y=400

Solution

semi−major axis b=5√6, semi−minor axis a=2√6

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25x²+4y²+100x−40y=400

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 25x²+4y²+100x−40y=400 in the form of the standard ellipse equation

(x−(−2))²(2√6)² +(y−5)²(5√6)² =1

Therefore ellipse properties are:

(h, k)=(−2, 5), a=2√6, b=5√6

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

Ellipse with center (h, k)=(−2, 5), b=5√6, a=2√6

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

trục 25x^2+4y^2+100x-40y=400

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