What is ((x+2)^2}{16}-\frac{(y+3)^2)/9 =1 ?

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

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

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

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

\frac{( y - 4 ) ^{2}}{48} - \frac{( x + 1 ) ^{2}}{9} = 1

25 y^{(2)} - 16 x^{(2)} - 400 = 0

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

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

x^{2} + 24 = y^{2} + 6 x - 2 y

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