What is the vertices (x^2)/(25)-(y^2)/(81)=1 ?

The vertices (x^2)/(25)-(y^2)/(81)=1 is (5,0),(-5,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices (x^2)/(25)-(y^2)/(81)=1

vertices

\frac{x ^{2}}{25} - \frac{y ^{2}}{81} = 1

(5, 0), (- 5, 0)

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{25} - \frac{y ^{2}}{81} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 5, b = 9

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

Refine

(5, 0), (- 5, 0)

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

+ \frac{( y - 2 ) ^{2}}{9} - \frac{( x - 1 ) ^{2}}{16} = 1

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

4 x^{2} - y^{2} - 16 x - 6 y - 9 = 0

4 x^{2} - 9 y^{2} = 144

What is the vertices (x^2)/(25)-(y^2)/(81)=1 ?
The vertices (x^2)/(25)-(y^2)/(81)=1 is (5,0),(-5,0)