center-16x^2+25y^2-64x-150y-239=0

Lösung

center

- 16 x^{2} + 25 y^{2} - 64 x - 150 y - 239 = 0

(- 2, 3)

Schritte zur Lösung

- 16 x^{2} + 25 y^{2} - 64 x - 150 y - 239 = 0

Rewrite

- 16 x^{2} + 25 y^{2} - 64 x - 150 y - 239 = 0

in the form of a standard hyperbola equation

\frac{( y - 3 ) ^{2}}{4 ^{2}} - \frac{( x - ( - 2 ) ) ^{2}}{5 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (- 2, 3), a = 4, b = 5

And the center is:

(- 2, 3)

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

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

ce n t er \frac{( y + 5 ) ^{2}}{9} - \frac{( x - 3 ) ^{2}}{25} = 9

ce n t er \frac{x ^{2}}{16} - \frac{y ^{2}}{9} = 1

ce n t er \frac{( y - 7 ) ^{2}}{9} - (x + 4)^{2} = 1