What is the foci-4x^2+24x+16y^2-128y+156=0 ?

The foci-4x^2+24x+16y^2-128y+156=0 is (3,4+2sqrt(5)),(3,4-2sqrt(5))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci-4x^2+24x+16y^2-128y+156=0

foci

- 4 x^{2} + 24 x + 16 y^{2} - 128 y + 156 = 0

(3, 4 + 25), (3, 4 - 25)

Solution steps

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

Calculate Hyperbola properties

- 4 x^{2} + 24 x + 16 y^{2} - 128 y + 156 = 0 :

Up-down Hyperbola with

(h, k) = (3, 4), a = 2, b = 4

(3, 4 + c), (3, 4 - c)

Compute

c :

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

(3, 4 + 25), (3, 4 - 25)

\frac{y ^{2}}{36} - \frac{x ^{2}}{16} = 1

\frac{( x + 1.6 ) ^{2}}{1} - \frac{( y + 1 ) ^{2}}{0.07} = 1

9 x^{2} - 4 y^{2} + 36 x + 8 y + 33 = 0

\frac{y ^{2}}{4} - \frac{x ^{2}}{25} = 1

25 y^{2} - 36 (x - 8)^{2} = 900

What is the foci-4x^2+24x+16y^2-128y+156=0 ?
The foci-4x^2+24x+16y^2-128y+156=0 is (3,4+2sqrt(5)),(3,4-2sqrt(5))