exzentrizität (x^2)/(324)+(y^2)/(400)=1

Lösung

exzentrizität

\frac{x ^{2}}{324} + \frac{y ^{2}}{400} = 1

\frac{19}{10}

Dezimale

0.43588 \dots

Schritte zur Lösung

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

Calculate ellipse properties

\frac{x ^{2}}{324} + \frac{y ^{2}}{400} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 20, a = 18

= \frac{2 0 ^{2} - 1 8 ^{2}}{20}

\frac{2 0 ^{2} - 1 8 ^{2}}{20} = \frac{19}{10}

= \frac{19}{10}

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

ecce n t r i c i t y 4 x^{2} + 3 y^{2} + 24 x + 36 y + 72 = 0

ecce n t r i c i t y \frac{5}{81} x^{2} + \frac{y ^{2}}{9} = 1

ecce n t r i c i t y 3 x^{2} + y^{2} - 24 x = - 39

ecce n t r i c i t y 9 x^{2} + 25 y^{2} + 18 x + 150 y + 9 = 0