What is the foci ((y-3)^2}{36}-\frac{(x+2)^2)/4 =1 ?

The foci ((y-3)^2}{36}-\frac{(x+2)^2)/4 =1 is (-2,3+2sqrt(10)),(-2,3-2sqrt(10))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci ((y-3)^2}{36}-\frac{(x+2)^2)/4 =1

foci

\frac{( y - 3 ) ^{2}}{36} - \frac{( x + 2 ) ^{2}}{4} = 1

(- 2, 3 + 210), (- 2, 3 - 210)

Solution steps

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

Calculate Hyperbola properties

\frac{( y - 3 ) ^{2}}{36} - \frac{( x + 2 ) ^{2}}{4} = 1 :

Up-down Hyperbola with

(h, k) = (- 2, 3), a = 6, b = 2

(- 2, 3 + c), (- 2, 3 - c)

Compute

c :

c = 6^{2} + 2^{2} : 210

(- 2, 3 + 210), (- 2, 3 - 210)

\frac{x ^{2}}{9} - \frac{y ^{2}}{55} = 1

\frac{( y + 3 ) ^{2}}{4} - \frac{( x + 1 ) ^{2}}{16} = 1

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

4 x^{2} - y^{2} = 16

- 9 x^{2} + y^{2} + 18 x - 45 = 0

What is the foci ((y-3)^2}{36}-\frac{(x+2)^2)/4 =1 ?
The foci ((y-3)^2}{36}-\frac{(x+2)^2)/4 =1 is (-2,3+2sqrt(10)),(-2,3-2sqrt(10))