What is the center 25x^2+4y^2-50x+16y-59=0 ?

The center 25x^2+4y^2-50x+16y-59=0 is (1,-2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

center 25x^2+4y^2-50x+16y-59=0

center

(1, - 2)

Solution steps

25 x^{2} + 4 y^{2} - 50 x + 16 y - 59 = 0

Rewrite

25 x^{2} + 4 y^{2} - 50 x + 16 y - 59 = 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (1, - 2), a = 2, b = 5

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (1, - 2), b = 5, a = 2

And the center is:

(1, - 2)

What is the center 25x^2+4y^2-50x+16y-59=0 ?
The center 25x^2+4y^2-50x+16y-59=0 is (1,-2)