What is the vertices 4(x-2)^2-9(y+4)^2=1 ?

The vertices 4(x-2)^2-9(y+4)^2=1 is (5/2 ,-4),(3/2 ,-4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices 4(x-2)^2-9(y+4)^2=1

vertices

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

(\frac{5}{2}, - 4), (\frac{3}{2}, - 4)

Solution steps

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

Calculate Hyperbola properties

4 (x - 2)^{2} - 9 (y + 4)^{2} = 1 :

Right-left Hyperbola with

(h, k) = (2, - 4), a = \frac{1}{2}, b = \frac{1}{3}

(2 + \frac{1}{2}, - 4), (2 - \frac{1}{2}, - 4)

Refine

(\frac{5}{2}, - 4), (\frac{3}{2}, - 4)

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

\frac{y ^{2}}{225} - \frac{x ^{2}}{64} = 1

\frac{y ^{2}}{1} - \frac{x ^{2}}{36} = 1

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

9 x^{2} - y^{2} + 18 x + 8 y = 43

What is the vertices 4(x-2)^2-9(y+4)^2=1 ?
The vertices 4(x-2)^2-9(y+4)^2=1 is (5/2 ,-4),(3/2 ,-4)