What is the vertices (x^2+8(y+5)^2)=20 ?

The vertices (x^2+8(y+5)^2)=20 is (2sqrt(5),-5),(-2sqrt(5),-5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices (x^2+8(y+5)^2)=20

vertices

(x^{2} + 8 (y + 5)^{2}) = 20

(25, - 5), (- 25, - 5)

Solution steps

(h + a, k), (h - a, k)

Calculate ellipse properties

(x^{2} + 8 (y + 5)^{2}) = 20 :

Ellipse with center

(h, k) = (0, - 5), a = 25, b = \frac{5}{2}

(0 + 25, - 5), (0 - 25, - 5)

Refine

(25, - 5), (- 25, - 5)

x^{2} + 4 y^{2} = 4 x + 8

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

\frac{( x - 3 ) ^{2}}{64} + \frac{( y + 5 ) ^{2}}{81} = 1

x^{2} + 4 y^{2} + 6 x + 16 y + 21 = 0

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

What is the vertices (x^2+8(y+5)^2)=20 ?
The vertices (x^2+8(y+5)^2)=20 is (2sqrt(5),-5),(-2sqrt(5),-5)