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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

f (x) = x^{2} - 2 x - 8

x^{2} - y = 4

16 x^{2} + y^{2} = 16

y = x^{2} - 6 x

x^{2} + 4 y^{2} - 6 x - 16 y + 21 = 0

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