What is the vertices (x^2)/(25)+((y+3)^2)/(49)=1 ?

The vertices (x^2)/(25)+((y+3)^2)/(49)=1 is (0,4),(0,-10)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices (x^2)/(25)+((y+3)^2)/(49)=1

vertices

\frac{x ^{2}}{25} + \frac{( y + 3 ) ^{2}}{49} = 1

(0, 4), (0, - 10)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{25} + \frac{( y + 3 ) ^{2}}{49} = 1 :

Ellipse with center

(h, k) = (0, - 3), b = 7, a = 5

(0, - 3 + 7), (0, - 3 - 7)

Refine

(0, 4), (0, - 10)

9 x^{2} + 25 y^{2} = 225

81 x^{2} - 1296 x + 16 y^{2} + 64 y = - 3952

6 x^{2} + 8 y^{2} + 12 x - 32 y = 178

49 x^{2} + 25 y^{2} - 196 x + 250 y - 404 = 0

\frac{( x - 3 ) ^{2}}{49} + \frac{( y + 5 ) ^{2}}{25} = 1

What is the vertices (x^2)/(25)+((y+3)^2)/(49)=1 ?
The vertices (x^2)/(25)+((y+3)^2)/(49)=1 is (0,4),(0,-10)