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

The vertices ((x+1)^2)/(25)+((y+1)^2)/(49)=1 is (-1,6),(-1,-8)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

vertices

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

(- 1, 6), (- 1, - 8)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (- 1, - 1), b = 7, a = 5

(- 1, - 1 + 7), (- 1, - 1 - 7)

Refine

(- 1, 6), (- 1, - 8)

\frac{( x - 1 ) ^{2}}{6} + \frac{( y + 2 ) ^{2}}{18} = 1

\frac{x ^{2}}{100} + \frac{y ^{2}}{52} = 1

x^{2} + 9 y^{2} + 16 x + 18 y + 64 = 0

x^{2} + 3 y^{2} + 6 x - 18 y + 33 = 0

\frac{( x + 1 ) ^{2}}{64} + \frac{( y - 6 ) ^{2}}{16} = 1

What is the vertices ((x+1)^2)/(25)+((y+1)^2)/(49)=1 ?
The vertices ((x+1)^2)/(25)+((y+1)^2)/(49)=1 is (-1,6),(-1,-8)