What is ((x-7)^2)/9-((y+3)^2)/(16)=1 ?

The solution to ((x-7)^2)/9-((y+3)^2)/(16)=1 is Hyperbola with (h,k)=(7,-3),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-7)^2)/9-((y+3)^2)/(16)=1

\frac{( x - 7 ) ^{2}}{9} - \frac{( y + 3 ) ^{2}}{16} = 1

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

Solution steps

\frac{( x - 7 ) ^{2}}{9} - \frac{( y + 3 ) ^{2}}{16} = 1

Rewrite

\frac{( x - 7 ) ^{2}}{9} - \frac{( y + 3 ) ^{2}}{16} = 1

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

144 x^{2} - 25 y^{2} + 864 x - 100 y + 2004 = 0

y^{2} - 4 x^{2} + 4 y + 24 x - 41 = 0

x, 3 m (x)^{2} - 2 x^{2} - 1 = 0

5 x^{2} - 4 y^{2} - 80 = 0

y + y^{2} = x^{2} + 4

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