2x^2+3y^2-4x+12y-20=0

Lösung

2 x^{2} + 3 y^{2} - 4 x + 12 y - 20 = 0

(h, k) = (1, - 2), a = 17, b = \frac{34}{3}

Schritte zur Lösung

2 x^{2} + 3 y^{2} - 4 x + 12 y - 20 = 0

Rewrite

2 x^{2} + 3 y^{2} - 4 x + 12 y - 20 = 0

in the form of the standard ellipse equation

\frac{( x - 1 ) ^{2}}{( 17 ) ^{2}} + \frac{( y - ( - 2 ) ) ^{2}}{( \frac{34}{3} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (1, - 2), a = 17, b = \frac{34}{3}

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (1, - 2), a = 17, b = \frac{34}{3}

\frac{( x - 3 ) ^{2}}{4} + \frac{( y - 3 ) ^{2}}{9} = 1

\frac{x ^{2}}{100} + \frac{y ^{2}}{4} = 1

2 x^{2} + 4 y^{2} + 9 x + 9 y - 17 = 0

x^{2} + 4 y^{2} - 2 x + 4 y + 1 = 0

16 x^{2} + 4 y^{2} + 96 x - 8 y + 84 = 0