What is 16x^2-160x-16y^2-128y-112=0 ?

The solution to 16x^2-160x-16y^2-128y-112=0 is Hyperbola with (h,k)=(5,-4),a=4,b=4

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16x^2-160x-16y^2-128y-112=0

16 x^{2} - 160 x - 16 y^{2} - 128 y - 112 = 0

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

Solution steps

16 x^{2} - 160 x - 16 y^{2} - 128 y - 112 = 0

Rewrite

16 x^{2} - 160 x - 16 y^{2} - 128 y - 112 = 0

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

[f (x)]^{2} = 3 x^{2} - x + 2

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

y^{2} = 4 + z^{2} - 2 z

a^{2} - b^{2} = 51

10 y^{2} - 9 x^{2} = 144

What is 16x^2-160x-16y^2-128y-112=0 ?
The solution to 16x^2-160x-16y^2-128y-112=0 is Hyperbola with (h,k)=(5,-4),a=4,b=4