((x-3)^2)/((18))+((y+1)^2)/(12)=1

Lösung

\frac{( x - 3 ) ^{2}}{( 18 )} + \frac{( y + 1 ) ^{2}}{12} = 1

(h, k) = (3, - 1), a = 32, b = 23

Schritte zur Lösung

\frac{( x - 3 ) ^{2}}{18} + \frac{( y + 1 ) ^{2}}{12} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{18} + \frac{( y + 1 ) ^{2}}{12} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (3, - 1), a = 32, b = 23

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (3, - 1), a = 32, b = 23

4 x^{2} + 8 y^{2} - 16 \cdot + 32 y + 32 = 0

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

490 x^{(2)} + 10 y^{(2)} + 808 x - 556 y + 40 = 0

\frac{x ^{2}}{2.3 2 ^{2}} + \frac{y ^{2}}{2.8 2 ^{2}} = 1

- 8 y^{2} + 72 = 9 x^{2}