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

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

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

(h, k) = (3, 2), a = 3, b = 4

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (3, 2), a = 3, b = 4

a^{2} - b^{2} = 2017

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

1 = aa - bb

25 x^{2} - 36 y^{2} - 150 x - 288 y - 1251 = 0

x^{2} + y^{2} + 5 x + y - 26 = 0.2 x^{2} + 2 y^{2} - 4 x + 9 y + 4

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