What is the foci ((x+2)^2)/(64)-((y-3)^2)/(36)=1 ?

The foci ((x+2)^2)/(64)-((y-3)^2)/(36)=1 is (8,3),(-12,3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci ((x+2)^2)/(64)-((y-3)^2)/(36)=1

foci

\frac{( x + 2 ) ^{2}}{64} - \frac{( y - 3 ) ^{2}}{36} = 1

(8, 3), (- 12, 3)

Solution steps

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

Calculate Hyperbola properties

\frac{( x + 2 ) ^{2}}{64} - \frac{( y - 3 ) ^{2}}{36} = 1 :

Right-left Hyperbola with

(h, k) = (- 2, 3), a = 8, b = 6

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

Compute

c :

c = 8^{2} + 6^{2} : 10

(- 2 + 10, 3), (- 2 - 10, 3)

Refine

(8, 3), (- 12, 3)

\frac{( y + 2 ) ^{2}}{25} - \frac{( x + 4 ) ^{2}}{9} = 1

(\frac{x - 6}{4})^{2} - (\frac{y}{4})^{2} = 1

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

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

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

What is the foci ((x+2)^2)/(64)-((y-3)^2)/(36)=1 ?
The foci ((x+2)^2)/(64)-((y-3)^2)/(36)=1 is (8,3),(-12,3)