What is the vertices ((x)^2)/(44)+(y^2)/(125)=1 ?

The vertices ((x)^2)/(44)+(y^2)/(125)=1 is (0,5sqrt(5)),(0,-5sqrt(5))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices ((x)^2)/(44)+(y^2)/(125)=1

vertices

\frac{( x ) ^{2}}{44} + \frac{y ^{2}}{125} = 1

(0, 55), (0, - 55)

Solution steps

(h, k + b), (h, k - b)

Calculate ellipse properties

\frac{x ^{2}}{44} + \frac{y ^{2}}{125} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 55, a = 211

(0, 0 + 55), (0, 0 - 55)

Refine

(0, 55), (0, - 55)

\frac{( x - 5 ) ^{2}}{36} + \frac{( y + 6 ) ^{2}}{25} = 1

81 x^{2} + y^{2} = 81

9 x^{2} + 4 y^{2} + 18 x - 8 y - 23 = 0

\frac{( x + 1 ) ^{2}}{16} + \frac{( y + 1 ) ^{2}}{25} = 1

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

What is the vertices ((x)^2)/(44)+(y^2)/(125)=1 ?
The vertices ((x)^2)/(44)+(y^2)/(125)=1 is (0,5sqrt(5)),(0,-5sqrt(5))