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

The solution to ((x+2)^2)/(10)-((y+4)^2)/(25)=1 is Hyperbola with (h,k)=(-2,-4),a=sqrt(10),b=5

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

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

(h, k) = (- 2, - 4), a = 10, b = 5

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

\frac{( x - ( - 2 ) ) ^{2}}{( 10 ) ^{2}} - \frac{( y - ( - 4 ) ) ^{2}}{5 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (- 2, - 4), a = 10, b = 5

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

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

9 x^{2} - 54 x + 9 - 4 y^{2} - 8 y - 113 = 0

y^{2} = 9 (1 + x^{2})

\frac{x ^{2}}{12} - \frac{y ^{2}}{16} = 1

What is ((x+2)^2)/(10)-((y+4)^2)/(25)=1 ?
The solution to ((x+2)^2)/(10)-((y+4)^2)/(25)=1 is Hyperbola with (h,k)=(-2,-4),a=sqrt(10),b=5