exzentrizität (x^2)/9+((y+5)^2)/(25)=1

Lösung

exzentrizität

\frac{x ^{2}}{9} + \frac{( y + 5 ) ^{2}}{25} = 1

\frac{4}{5}

Dezimale

0.8

Schritte zur Lösung

= \frac{b ^{2} - a ^{2}}{b}

Calculate ellipse properties

\frac{x ^{2}}{9} + \frac{( y + 5 ) ^{2}}{25} = 1 :

Ellipse with center

(h, k) = (0, - 5), b = 5, a = 3

= \frac{5 ^{2} - 3 ^{2}}{5}

\frac{5 ^{2} - 3 ^{2}}{5} = \frac{4}{5}

= \frac{4}{5}

ecce n t r i c i t y 4 y^{2} + 9 x^{2} = 36

ecce n t r i c i t y 2 x^{2} + 3 y^{2} + 4 x = 1

ecce n t r i c i t y 25 x^{2} + 16 y^{2} + 100 x - 96 y - 156 = 0

ecce n t r i c i t y 18 x^{2} + 8 y^{2} + 36 x - 32 y - 22 = 0

ecce n t r i c i t y \frac{x ^{2}}{5} + \frac{y ^{2}}{4.58} = 1