What is the vertices ((x-2)^2)/9+((y+3)^2)/5 =1 ?

The vertices ((x-2)^2)/9+((y+3)^2)/5 =1 is (5,-3),(-1,-3)

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+3)^2)/5 =1

vertices

\frac{( x - 2 ) ^{2}}{9} + \frac{( y + 3 ) ^{2}}{5} = 1

(5, - 3), (- 1, - 3)

Solution steps

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

Calculate ellipse properties

\frac{( x - 2 ) ^{2}}{9} + \frac{( y + 3 ) ^{2}}{5} = 1 :

Ellipse with center

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

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

Refine

(5, - 3), (- 1, - 3)

2 x^{2} + 3 y^{2} + 4 x - 1 = 0

25 x^{2} + 9 y^{2} + 200 x - 90 y + 400 = 0

9 x^{2} + 36 y^{2} + 36 x - 216 y + 36 = 0

9 x^{2} + 6 y^{2} + 36 y = 0

\frac{x ^{2}}{2} + \frac{y ^{2}}{3} = 1

What is the vertices ((x-2)^2)/9+((y+3)^2)/5 =1 ?
The vertices ((x-2)^2)/9+((y+3)^2)/5 =1 is (5,-3),(-1,-3)