What is the vertices ((x-2)^2)/9+((y+5)^2)/4 =1 ?

The vertices ((x-2)^2)/9+((y+5)^2)/4 =1 is (5,-5),(-1,-5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices ((x-2)^2)/9+((y+5)^2)/4 =1

vertices

\frac{( x - 2 ) ^{2}}{9} + \frac{( y + 5 ) ^{2}}{4} = 1

(5, - 5), (- 1, - 5)

Solution steps

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

Calculate ellipse properties

\frac{( x - 2 ) ^{2}}{9} + \frac{( y + 5 ) ^{2}}{4} = 1 :

Ellipse with center

(h, k) = (2, - 5), a = 3, b = 2

(2 + 3, - 5), (2 - 3, - 5)

Refine

(5, - 5), (- 1, - 5)

\frac{( x - 4 ) ^{2}}{24} + \frac{( y - 2 ) ^{2}}{49} = 1

25 x^{2} + 49 y^{2} = 1225

\frac{2 ( x - 6 ) ^{2}}{3} + \frac{7 ( y + 4 ) ^{2}}{4} = 1 x

\frac{x ^{2}}{12} + \frac{y ^{2}}{144} = \frac{y}{12}

81 x^{2} + 36 y^{2} = 2916

What is the vertices ((x-2)^2)/9+((y+5)^2)/4 =1 ?
The vertices ((x-2)^2)/9+((y+5)^2)/4 =1 is (5,-5),(-1,-5)