exzentrizität 25x^2+16y^2=400

Lösung

exzentrizität

25 x^{2} + 16 y^{2} = 400

\frac{3}{5}

Dezimale

0.6

Schritte zur Lösung

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

Calculate ellipse properties

25 x^{2} + 16 y^{2} = 400 :

Ellipse with center

(h, k) = (0, 0), b = 5, a = 4

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

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

= \frac{3}{5}

ecce n t r i c i t y x^{2} + 4 y^{2} - 6 x - 16 y + 21 = 0

ecce n t r i c i t y \frac{x ^{2}}{16} + \frac{y ^{2}}{12} = 1

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

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

ecce n t r i c i t y \frac{x ^{2}}{64} + \frac{y ^{2}}{100} = 1