What is ((x+2)/(10))^2-((y+4)/(25))^2=1 ?

The solution to ((x+2)/(10))^2-((y+4)/(25))^2=1 is Hyperbola with (h,k)=(-2,-4),a=10,b=25

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((x+2)/(10))^2-((y+4)/(25))^2=1

(\frac{x + 2}{10})^{2} - (\frac{y + 4}{25})^{2} = 1

(h, k) = (- 2, - 4), a = 10, b = 25

Solution steps

(\frac{x + 2}{10})^{2} - (\frac{y + 4}{25})^{2} = 1

Rewrite

(\frac{x + 2}{10})^{2} - (\frac{y + 4}{25})^{2} = 1

in the form of a standard hyperbola equation

\frac{( x - ( - 2 ) ) ^{2}}{1 0 ^{2}} - \frac{( y - ( - 4 ) ) ^{2}}{2 5 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (- 2, - 4), a = 10, b = 25

4 (x + 1)^{2} - 3 (y - 2)^{2} = 15

4 x^{2} - 3 y^{2} + 8 x + 16 = 0

(x + 5)^{2} - (y + 6)^{2} = 20

- x^{2} + 2 y^{2} = 1

\frac{x ^{2}}{100} - \frac{y ^{2}}{75} = 1

What is ((x+2)/(10))^2-((y+4)/(25))^2=1 ?
The solution to ((x+2)/(10))^2-((y+4)/(25))^2=1 is Hyperbola with (h,k)=(-2,-4),a=10,b=25