((x+3)^2)/4+((y-4)^2)/(18)=1

Lösung

\frac{( x + 3 ) ^{2}}{4} + \frac{( y - 4 ) ^{2}}{18} = 1

(h, k) = (- 3, 4), a = 2, b = 32

Schritte zur Lösung

\frac{( x + 3 ) ^{2}}{4} + \frac{( y - 4 ) ^{2}}{18} = 1

Rewrite

\frac{( x + 3 ) ^{2}}{4} + \frac{( y - 4 ) ^{2}}{18} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 3, 4), a = 2, b = 32

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 3, 4), b = 32, a = 2

\frac{( x + 1 ) ^{2}}{\frac{27}{4}} + \frac{( y - 4 ) ^{2}}{9} = 1

4 x^{2} + 9 y^{2} - 8 x - 54 y - 5 = 0

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

9 x^{2} + 16 y^{2} - 160 y + 256 = 0

\frac{x ^{2}}{10} + \frac{y ^{2}}{2} = 1