What is ((y-3)^2)/(49)-((x+2)^2)/(16)=1 ?

The solution to ((y-3)^2)/(49)-((x+2)^2)/(16)=1 is Hyperbola with (h,k)=(-2,3),a=7,b=4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

((y-3)^2)/(49)-((x+2)^2)/(16)=1

\frac{( y - 3 ) ^{2}}{49} - \frac{( x + 2 ) ^{2}}{16} = 1

(h, k) = (- 2, 3), a = 7, b = 4

Solution steps

\frac{( y - 3 ) ^{2}}{49} - \frac{( x + 2 ) ^{2}}{16} = 1

Rewrite

\frac{( y - 3 ) ^{2}}{49} - \frac{( x + 2 ) ^{2}}{16} = 1

in the form of a standard hyperbola equation

\frac{( y - 3 ) ^{2}}{7 ^{2}} - \frac{( x - ( - 2 ) ) ^{2}}{4 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (- 2, 3), a = 7, b = 4

y^{2} \leq x^{2} + y

36 y^{2} = 4 x^{2} + 144

y^{2} - 9 = x^{2}

3 y^{2} - 12 x^{2} - 36 y - 48 x + 48 = 0

\frac{( y + 7 ) ^{2}}{4} - \frac{( x + 2 ) ^{2}}{16} = 1

What is ((y-3)^2)/(49)-((x+2)^2)/(16)=1 ?
The solution to ((y-3)^2)/(49)-((x+2)^2)/(16)=1 is Hyperbola with (h,k)=(-2,3),a=7,b=4