What is the foci ((x-5)^2)/(16)-((y-4)^2)/(64)=1 ?

The foci ((x-5)^2)/(16)-((y-4)^2)/(64)=1 is (5+4sqrt(5),4),(5-4sqrt(5),4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci ((x-5)^2)/(16)-((y-4)^2)/(64)=1

foci

\frac{( x - 5 ) ^{2}}{16} - \frac{( y - 4 ) ^{2}}{64} = 1

(5 + 45, 4), (5 - 45, 4)

Solution steps

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

Calculate Hyperbola properties

\frac{( x - 5 ) ^{2}}{16} - \frac{( y - 4 ) ^{2}}{64} = 1 :

Right-left Hyperbola with

(h, k) = (5, 4), a = 4, b = 8

(5 + c, 4), (5 - c, 4)

Compute

c :

c = 4^{2} + 8^{2} : 45

(5 + 45, 4), (5 - 45, 4)

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

4 x^{2} - 12 y^{2} - 16 x - 72 y - 44 = 0

- 25 x^{2} + 36 y^{2} = 900

x^{2} - 9 y^{2} = 36

\frac{( y + 5 ) ^{2}}{9} - \frac{( x + 2 ) ^{2}}{16} = 1

What is the foci ((x-5)^2)/(16)-((y-4)^2)/(64)=1 ?
The foci ((x-5)^2)/(16)-((y-4)^2)/(64)=1 is (5+4sqrt(5),4),(5-4sqrt(5),4)