What is the vertices 9x^2+16y^2+36x-160y+292=0 ?

The vertices 9x^2+16y^2+36x-160y+292=0 is (2,5),(-6,5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices 9x^2+16y^2+36x-160y+292=0

vertices

9 x^{2} + 16 y^{2} + 36 x - 160 y + 292 = 0

(2, 5), (- 6, 5)

Solution steps

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

Calculate ellipse properties

9 x^{2} + 16 y^{2} + 36 x - 160 y + 292 = 0 :

Ellipse with center

(h, k) = (- 2, 5), a = 4, b = 3

(- 2 + 4, 5), (- 2 - 4, 5)

Refine

(2, 5), (- 6, 5)

\frac{x ^{2}}{36} + \frac{y ^{2}}{81} = 1

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

27 x^{2} + y^{2} - 18 y = 0

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

25 x^{2} + y^{2} - 4 y - 21 = 0

What is the vertices 9x^2+16y^2+36x-160y+292=0 ?
The vertices 9x^2+16y^2+36x-160y+292=0 is (2,5),(-6,5)