What is the foci ((y+3)^2}{16}-\frac{(x+1)^2)/9 =1 ?

The foci ((y+3)^2}{16}-\frac{(x+1)^2)/9 =1 is (-1,2),(-1,-8)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci ((y+3)^2}{16}-\frac{(x+1)^2)/9 =1

foci

\frac{( y + 3 ) ^{2}}{16} - \frac{( x + 1 ) ^{2}}{9} = 1

(- 1, 2), (- 1, - 8)

Solution steps

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

Calculate Hyperbola properties

\frac{( y + 3 ) ^{2}}{16} - \frac{( x + 1 ) ^{2}}{9} = 1 :

Up-down Hyperbola with

(h, k) = (- 1, - 3), a = 4, b = 3

(- 1, - 3 + c), (- 1, - 3 - c)

Compute

c :

c = 4^{2} + 3^{2} : 5

(- 1, - 3 + 5), (- 1, - 3 - 5)

Refine

(- 1, 2), (- 1, - 8)

4 x^{2} - 9 y^{2} + 3 x + 18 y + 91 = 0

4 x^{2} - 25 y^{2} = 100

\frac{y ^{2}}{3} - \frac{x ^{2}}{3} = 1

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

\frac{( y + 3 ) ^{2}}{10} - \frac{( x - 3 ) ^{2}}{20} = 1

What is the foci ((y+3)^2}{16}-\frac{(x+1)^2)/9 =1 ?
The foci ((y+3)^2}{16}-\frac{(x+1)^2)/9 =1 is (-1,2),(-1,-8)