5x^2-3y^2+31x+23y+24=0

Lösung

5 x^{2} - 3 y^{2} + 31 x + 23 y + 24 = 0

(h, k) = (- \frac{31}{10}, \frac{23}{6}), a = \frac{601}{3 10}, b = \frac{601}{5 6}

Schritte zur Lösung

5 x^{2} - 3 y^{2} + 31 x + 23 y + 24 = 0

Rewrite

5 x^{2} - 3 y^{2} + 31 x + 23 y + 24 = 0

in the form of a standard hyperbola equation

\frac{( y - \frac{23}{6} ) ^{2}}{( \frac{601}{3 10} ) ^{2}} - \frac{( x - ( - \frac{31}{10} ) ) ^{2}}{( \frac{601}{5 6} ) ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (- \frac{31}{10}, \frac{23}{6}), a = \frac{601}{3 10}, b = \frac{601}{5 6}

y^{2} = \frac{( 900 + 36 ( x - 8 ) ^{2} )}{25}

\frac{x ^{2}}{400} - \frac{y ^{2}}{625} = 1

3 x^{2} - 9 y^{2} - 18 x + 18 y - 54 = 0

p^{2} = 32 + q^{2}

\frac{( x - 3 ) ^{2}}{( 3 ) ^{2}} - \frac{( y + 5 ) ^{2}}{( 6 ) ^{2}} = 1