What is ((x-3)^2)/(25)-((y-2)^2)/(40)=1 ?

The solution to ((x-3)^2)/(25)-((y-2)^2)/(40)=1 is Hyperbola with (h,k)=(3,2),a=5,b=2sqrt(10)

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((x-3)^2)/(25)-((y-2)^2)/(40)=1

\frac{( x - 3 ) ^{2}}{25} - \frac{( y - 2 ) ^{2}}{40} = 1

(h, k) = (3, 2), a = 5, b = 210

Solution steps

\frac{( x - 3 ) ^{2}}{25} - \frac{( y - 2 ) ^{2}}{40} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{25} - \frac{( y - 2 ) ^{2}}{40} = 1

in the form of a standard hyperbola equation

\frac{( x - 3 ) ^{2}}{5 ^{2}} - \frac{( y - 2 ) ^{2}}{( 2 10 ) ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (3, 2), a = 5, b = 210

\frac{( x - 1 ) ^{2}}{4} - \frac{( y + 2 ) ^{2}}{13} = 1

10 0^{2} + 15 t^{2} = d^{2}

x^{2} - y^{2} - 6 x - 2 y - 10 = 0

y^{2} = x^{2} - 8 x

X^{2} - 81 Y^{2} = 81

What is ((x-3)^2)/(25)-((y-2)^2)/(40)=1 ?
The solution to ((x-3)^2)/(25)-((y-2)^2)/(40)=1 is Hyperbola with (h,k)=(3,2),a=5,b=2sqrt(10)