exzentrizität (x^2)/5+(y^2)/(20.98)=1

Lösung

exzentrizität

\frac{x ^{2}}{5} + \frac{y ^{2}}{20.98} = 1

0.87274 \dots

Schritte zur Lösung

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

Calculate ellipse properties

\frac{x ^{2}}{5} + \frac{y ^{2}}{20.98} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 20.98, a = 5

= \frac{( 20.98 ) ^{2} - ( 5 ) ^{2}}{20.98}

\frac{( 20.98 ) ^{2} - ( 5 ) ^{2}}{20.98} = 0.87274 \dots

= 0.87274 \dots

ecce n t r i c i t y 16 x^{2} + 192 x + 576 + 25 y^{2} = 1600

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

ecce n t r i c i t y 4 x^{2} + 9 y^{2} = 144

ecce n t r i c i t y \frac{x ^{2}}{16} + \frac{y ^{2}}{81} = 1

ecce n t r i c i t y \frac{x ^{2}}{5} + \frac{y ^{2}}{8} = 1