What is ((x-5)^2}{16}-\frac{(y+7)^2)/9 =-1 ?

The solution to ((x-5)^2}{16}-\frac{(y+7)^2)/9 =-1 is Hyperbola with (h,k)=(5,-7),a=3,b=4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

((x-5)^2}{16}-\frac{(y+7)^2)/9 =-1

\frac{( x - 5 ) ^{2}}{16} - \frac{( y + 7 ) ^{2}}{9} = - 1

(h, k) = (5, - 7), a = 3, b = 4

Solution steps

\frac{( x - 5 ) ^{2}}{16} - \frac{( y + 7 ) ^{2}}{9} = - 1

Rewrite

\frac{( x - 5 ) ^{2}}{16} - \frac{( y + 7 ) ^{2}}{9} = - 1

in the form of a standard hyperbola equation

\frac{( y - ( - 7 ) ) ^{2}}{3 ^{2}} - \frac{( x - 5 ) ^{2}}{4 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (5, - 7), a = 3, b = 4

2 x^{2} - y^{2} - 1 = 0

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

\frac{( x - 3 ) ^{2}}{49} - \frac{( y + 5 ) ^{2}}{25} = 1

\frac{( x - 8 ) ^{2}}{9} - \frac{( y + 3 ) ^{2}}{16} = 1

\frac{x ^{2}}{9 + y ^{2}} = 1

What is ((x-5)^2}{16}-\frac{(y+7)^2)/9 =-1 ?
The solution to ((x-5)^2}{16}-\frac{(y+7)^2)/9 =-1 is Hyperbola with (h,k)=(5,-7),a=3,b=4