What is the foci 4x^2-25y^2+16x+50y-109=0 ?

The foci 4x^2-25y^2+16x+50y-109=0 is (-2+sqrt(29),1),(-2-sqrt(29),1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci 4x^2-25y^2+16x+50y-109=0

foci

4 x^{2} - 25 y^{2} + 16 x + 50 y - 109 = 0

(- 2 + 29, 1), (- 2 - 29, 1)

Solution steps

(h + c, k), (h - c, k)

Calculate Hyperbola properties

4 x^{2} - 25 y^{2} + 16 x + 50 y - 109 = 0 :

Right-left Hyperbola with

(h, k) = (- 2, 1), a = 5, b = 2

(- 2 + c, 1), (- 2 - c, 1)

Compute

c :

c = 5^{2} + 2^{2} : 29

(- 2 + 29, 1), (- 2 - 29, 1)

9 x^{2} - 18 x - 24 y - 63 = 4 y^{2}

36 x^{2} - 25 y^{2} = 900

\frac{x ^{2}}{36} - \frac{y ^{2}}{4} = 1

- 9 x^{2} - 32 y = - 16 y^{2} + 128

\frac{x ^{2}}{16} - \frac{y ^{2}}{16} = 1

What is the foci 4x^2-25y^2+16x+50y-109=0 ?
The foci 4x^2-25y^2+16x+50y-109=0 is (-2+sqrt(29),1),(-2-sqrt(29),1)