What is the center 25x^2+16y^2=400 ?

The center 25x^2+16y^2=400 is (0,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

center 25x^2+16y^2=400

center

25 x^{2} + 16 y^{2} = 400

(0, 0)

Solution steps

25 x^{2} + 16 y^{2} = 400

Rewrite

25 x^{2} + 16 y^{2} = 400

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{4 ^{2}} + \frac{( y - 0 ) ^{2}}{5 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 0), a = 4, b = 5

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (0, 0), b = 5, a = 4

And the center is:

(0, 0)

x^{2} + y^{2} - 10 y = 0

\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1

\frac{x ^{2}}{16} + \frac{y ^{2}}{9} = 1

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

y^{2} = 2 (4 x - 3 - x^{2})