exzentrizität 3x^2+12x+4y^2-40y+100=0

Lösung

exzentrizität

3 x^{2} + 12 x + 4 y^{2} - 40 y + 100 = 0

\frac{1}{2}

Dezimale

0.5

Schritte zur Lösung

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

Calculate ellipse properties

3 x^{2} + 12 x + 4 y^{2} - 40 y + 100 = 0 :

Ellipse with center

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

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

\frac{2 ^{2} - ( 3 ) ^{2}}{2} = \frac{1}{2}

= \frac{1}{2}

ecce n t r i c i t y 4 x^{2} + 9 y^{2} + 32 x - 18 y + 37 = 0

ecce n t r i c i t y 5 x^{2} + 7 y^{2} = 70

ecce n t r i c i t y 5 x^{2} + 10 y^{2} = 50

ecce n t r i c i t y x^{2} + 4 y^{2} - 2 x - 16 y + 13 = 0

ecce n t r i c i t y \frac{( x - 1 ) ^{2}}{6} + \frac{( y - 3 ) ^{2}}{2} = 1