exzentrizität 4x^2+25y^2-40x+150y+225=0

Lösung

exzentrizität

4 x^{2} + 25 y^{2} - 40 x + 150 y + 225 = 0

\frac{21}{5}

Dezimale

0.91651 \dots

Schritte zur Lösung

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

Calculate ellipse properties

4 x^{2} + 25 y^{2} - 40 x + 150 y + 225 = 0 :

Ellipse with center

(h, k) = (5, - 3), a = 5, b = 2

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

\frac{5 ^{2} - 2 ^{2}}{5} = \frac{21}{5}

= \frac{21}{5}

ecce n t r i c i t y \frac{x ^{2}}{9} + \frac{y ^{2}}{49} = 1

ecce n t r i c i t y \frac{x ^{2}}{100} + \frac{y ^{2}}{36} = 1

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

ecce n t r i c i t y \frac{x ^{2}}{16} + (y - 4)^{2} = 1

ecce n t r i c i t y 9 x^{2} + 4 y^{2} - 18 x + 8 y - 23 = 0