What is the vertices 16y^2-9x^2+36x+96y-36=0 ?

The vertices 16y^2-9x^2+36x+96y-36=0 is (2,0),(2,-6)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices 16y^2-9x^2+36x+96y-36=0

vertices

16 y^{2} - 9 x^{2} + 36 x + 96 y - 36 = 0

(2, 0), (2, - 6)

Solution steps

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

Calculate Hyperbola properties

16 y^{2} - 9 x^{2} + 36 x + 96 y - 36 = 0 :

Up-down Hyperbola with

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

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

Refine

(2, 0), (2, - 6)

- 25 x^{2} + 4 y^{2} = 100

\frac{y ^{2}}{144} - \frac{x ^{2}}{25} = 1

\frac{( x - 3 ) ^{2}}{9} - \frac{( y - 2 ) ^{2}}{4} = 1

\frac{( x + 1 ) ^{2}}{9} - \frac{( y - 1 ) ^{2}}{8} = 1

\frac{x ^{2}}{60} - \frac{y ^{2}}{40} = 1

What is the vertices 16y^2-9x^2+36x+96y-36=0 ?
The vertices 16y^2-9x^2+36x+96y-36=0 is (2,0),(2,-6)