What is ((x-4)^2)/(36)-((y+3)^2)/(25)=1 ?

The solution to ((x-4)^2)/(36)-((y+3)^2)/(25)=1 is Hyperbola with (h,k)=(4,-3),a=6,b=5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

((x-4)^2)/(36)-((y+3)^2)/(25)=1

\frac{( x - 4 ) ^{2}}{36} - \frac{( y + 3 ) ^{2}}{25} = 1

(h, k) = (4, - 3), a = 6, b = 5

Solution steps

\frac{( x - 4 ) ^{2}}{36} - \frac{( y + 3 ) ^{2}}{25} = 1

Rewrite

\frac{( x - 4 ) ^{2}}{36} - \frac{( y + 3 ) ^{2}}{25} = 1

in the form of a standard hyperbola equation

\frac{( x - 4 ) ^{2}}{6 ^{2}} - \frac{( y - ( - 3 ) ) ^{2}}{5 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (4, - 3), a = 6, b = 5

x^{2} - y^{2} = 36

x^{2} - 4 y^{2} + 6 x + 8 y = - 1

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

\frac{x ^{2}}{4 + y ^{2}} = 1

25 y^{2} - 4 x^{2} = 100

What is ((x-4)^2)/(36)-((y+3)^2)/(25)=1 ?
The solution to ((x-4)^2)/(36)-((y+3)^2)/(25)=1 is Hyperbola with (h,k)=(4,-3),a=6,b=5