What is ((x-3)^2)/(100)-((y+4)^2)/(169)=1 ?

The solution to ((x-3)^2)/(100)-((y+4)^2)/(169)=1 is Hyperbola with (h,k)=(3,-4),a=10,b=13

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((x-3)^2)/(100)-((y+4)^2)/(169)=1

\frac{( x - 3 ) ^{2}}{100} - \frac{( y + 4 ) ^{2}}{169} = 1

(h, k) = (3, - 4), a = 10, b = 13

Solution steps

\frac{( x - 3 ) ^{2}}{100} - \frac{( y + 4 ) ^{2}}{169} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{100} - \frac{( y + 4 ) ^{2}}{169} = 1

in the form of a standard hyperbola equation

\frac{( x - 3 ) ^{2}}{1 0 ^{2}} - \frac{( y - ( - 4 ) ) ^{2}}{1 3 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (3, - 4), a = 10, b = 13

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

\frac{x ^{2}}{9} - \frac{( y + 2 ) ^{2}}{4} = 1

x^{2} - z^{2} + 24 = 0

9 x^{2} - 4 y^{2} - 18 x - 16 y - 43 = 0

\frac{( x ^{2} )}{4} - \frac{( y ^{2} )}{9} = 1

What is ((x-3)^2)/(100)-((y+4)^2)/(169)=1 ?
The solution to ((x-3)^2)/(100)-((y+4)^2)/(169)=1 is Hyperbola with (h,k)=(3,-4),a=10,b=13