((x-2)^2)/(12)-((3y^2+y))/(12)=1

Lösung

\frac{( x - 2 ) ^{2}}{12} - \frac{( 3 y ^{2} + y )}{12} = 1

(h, k) = (2, - \frac{1}{6}), a = \frac{143}{2 3}, b = \frac{143}{6}

Schritte zur Lösung

\frac{( x - 2 ) ^{2}}{12} - \frac{3 y ^{2} + y}{12} = 1

Rewrite

\frac{( x - 2 ) ^{2}}{12} - \frac{3 y ^{2} + y}{12} = 1

in the form of a standard hyperbola equation

\frac{( x - 2 ) ^{2}}{( \frac{143}{2 3} ) ^{2}} - \frac{( y - ( - \frac{1}{6} ) ) ^{2}}{( \frac{143}{6} ) ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (2, - \frac{1}{6}), a = \frac{143}{2 3}, b = \frac{143}{6}

2 x - 4 x^{2} = - 9 y^{2} - 21 y - 50

\frac{( x + 3 ) ^{2}}{9} - \frac{( y - 1 ) ^{2}}{4} = 1

(\frac{K ( x )}{2})^{2} = 4096 + 4096 - 128 x + x^{2}

1 \cdot 3 = (a + b) (a - b)

x^{2} - y^{2} = - (1)