What is the vertices ((x-1)^2)/6+((y+2)^2)/(18)=1 ?

The vertices ((x-1)^2)/6+((y+2)^2)/(18)=1 is (1,-2+3sqrt(2)),(1,-2-3sqrt(2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices ((x-1)^2)/6+((y+2)^2)/(18)=1

vertices

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

(1, - 2 + 32), (1, - 2 - 32)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (1, - 2), b = 32, a = 6

(1, - 2 + 32), (1, - 2 - 32)

\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

\frac{x ^{2}}{100} + \frac{y ^{2}}{25} = 1

What is the vertices ((x-1)^2)/6+((y+2)^2)/(18)=1 ?
The vertices ((x-1)^2)/6+((y+2)^2)/(18)=1 is (1,-2+3sqrt(2)),(1,-2-3sqrt(2))