What is the foci ((x-4)^2}{49}-\frac{(y+6)^2)/9 =1 ?

The foci ((x-4)^2}{49}-\frac{(y+6)^2)/9 =1 is (4+sqrt(58),-6),(4-sqrt(58),-6)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci ((x-4)^2}{49}-\frac{(y+6)^2)/9 =1

foci

\frac{( x - 4 ) ^{2}}{49} - \frac{( y + 6 ) ^{2}}{9} = 1

(4 + 58, - 6), (4 - 58, - 6)

Solution steps

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

Calculate Hyperbola properties

\frac{( x - 4 ) ^{2}}{49} - \frac{( y + 6 ) ^{2}}{9} = 1 :

Right-left Hyperbola with

(h, k) = (4, - 6), a = 7, b = 3

(4 + c, - 6), (4 - c, - 6)

Compute

c :

c = 7^{2} + 3^{2} : 58

(4 + 58, - 6), (4 - 58, - 6)

\frac{( x ^{2} )}{36} - \frac{( y ^{2} )}{25} = 1

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

y^{2} - 4 x^{2} < 1

x^{2} - 16 y^{2} = 64

(\frac{( x ^{2} )}{27}) - \frac{( ( y ) ^{2} )}{36} = 1

What is the foci ((x-4)^2}{49}-\frac{(y+6)^2)/9 =1 ?
The foci ((x-4)^2}{49}-\frac{(y+6)^2)/9 =1 is (4+sqrt(58),-6),(4-sqrt(58),-6)