What is the vertices ((y-2)^2}{25}-\frac{x^2)/9 =1 ?

The vertices ((y-2)^2}{25}-\frac{x^2)/9 =1 is (0,7),(0,-3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices ((y-2)^2}{25}-\frac{x^2)/9 =1

vertices

\frac{( y - 2 ) ^{2}}{25} - \frac{x ^{2}}{9} = 1

(0, 7), (0, - 3)

Solution steps

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

Calculate Hyperbola properties

\frac{( y - 2 ) ^{2}}{25} - \frac{x ^{2}}{9} = 1 :

Up-down Hyperbola with

(h, k) = (0, 2), a = 5, b = 3

(0, 2 + 5), (0, 2 - 5)

Refine

(0, 7), (0, - 3)

- 4 x^{2} + y^{2} + 8 y - 20 = 0

\frac{y ^{2}}{25} - \frac{x ^{2}}{39} = 1

y^{2} - 9 x^{2} - 8 y - 54 x - 81 = 0

9 x^{2} - 16 y^{2} - 18 x + 64 y - 199 = 0

\frac{x ^{2}}{25} - \frac{y ^{2}}{49} = 1

What is the vertices ((y-2)^2}{25}-\frac{x^2)/9 =1 ?
The vertices ((y-2)^2}{25}-\frac{x^2)/9 =1 is (0,7),(0,-3)