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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

(h, k) = (- 2, 1), a = 4, b = 3

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (- 2, 1), a = 4, b = 3

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

(\frac{x + 1}{25})^{2} - (\frac{y - 4}{144})^{2} = 1

\frac{x ^{2}}{64} - \frac{y ^{2}}{100} = 1

y^{2} = (x - 2)^{2} - 1

y^{2} - x^{2} = 100

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