exzentrizität (x^2)/3+(y^2)/5 =1

Lösung

exzentrizität

\frac{x ^{2}}{3} + \frac{y ^{2}}{5} = 1

\frac{2}{5}

Dezimale

0.63245 \dots

Schritte zur Lösung

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

Calculate ellipse properties

\frac{x ^{2}}{3} + \frac{y ^{2}}{5} = 1 :

Ellipse with center

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

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

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

= \frac{2}{5}

ecce n t r i c i t y \frac{x ^{2}}{16} + \frac{y ^{2}}{9} = 1

ecce n t r i c i t y 14 x^{2} + 9 y^{2} = 126

ecce n t r i c i t y \frac{y ^{2}}{64} + \frac{x ^{2}}{36} = 1

ecce n t r i c i t y 3 x^{2} + y^{2} + 2 x - 6 y = 3

ecce n t r i c i t y 3 x^{2} + 12 x + 4 y^{2} - 40 y + 100 = 0