center 18x^2-32y^2-108y-128x-254=0^2

Lösung

center

18 x^{2} - 32 y^{2} - 108 y - 128 x - 254 = 0^{2}

(\frac{32}{9}, - \frac{27}{16})

Schritte zur Lösung

18 x^{2} - 32 y^{2} - 108 y - 128 x - 254 = 0^{2}

Rewrite

18 x^{2} - 32 y^{2} - 108 y - 128 x - 254 = 0^{2}

in the form of a standard hyperbola equation

\frac{( x - \frac{32}{9} ) ^{2}}{( \frac{28111}{36} ) ^{2}} - \frac{( y - ( - \frac{27}{16} ) ) ^{2}}{( \frac{28111}{48} ) ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (\frac{32}{9}, - \frac{27}{16}), a = \frac{28111}{36}, b = \frac{28111}{48}

And the center is:

(\frac{32}{9}, - \frac{27}{16})

ce n t er \frac{( x + 1 ) ^{2}}{36} - \frac{( y - 3 ) ^{2}}{25} = 1

ce n t er 10 x^{2} - 25 y^{2} = 100

ce n t er x^{2} - y^{2} - 8 x - 2 y - 19 = 0

ce n t er \frac{( x - 5 ) ^{2}}{25} - \frac{( y + 4 ) ^{2}}{9} = 1

ce n t er \frac{( x - 5 ) ^{2}}{7} - \frac{( y + 3 ) ^{2}}{9} = 1