What is the foci x^2-4y^2+16y-32=0 ?

The foci x^2-4y^2+16y-32=0 is (2sqrt(5),2),(-2sqrt(5),2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci x^2-4y^2+16y-32=0

foci

x^{2} - 4 y^{2} + 16 y - 32 = 0

(25, 2), (- 25, 2)

Solution steps

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

Calculate Hyperbola properties

x^{2} - 4 y^{2} + 16 y - 32 = 0 :

Right-left Hyperbola with

(h, k) = (0, 2), a = 4, b = 2

(0 + c, 2), (0 - c, 2)

Compute

c :

c = 4^{2} + 2^{2} : 25

(0 + 25, 2), (0 - 25, 2)

Refine

(25, 2), (- 25, 2)

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

4 x^{2} - 9 y^{2} - 8 x + 72 y - 176 = 0

- x^{2} + 4 y^{2} - 8 y - 12 = 0

3 y^{2} - 12 x^{2} - 36 y - 48 x + 48 = 0

\frac{y ^{2}}{16} - \frac{x ^{2}}{20} = 1

What is the foci x^2-4y^2+16y-32=0 ?
The foci x^2-4y^2+16y-32=0 is (2sqrt(5),2),(-2sqrt(5),2)