((2x+1)^2)/4+((3y-2)^2)/(16)=1

Lösung

\frac{( 2 x + 1 ) ^{2}}{4} + \frac{( 3 y - 2 ) ^{2}}{16} = 1

(h, k) = (- \frac{1}{2}, \frac{2}{3}), a = 1, b = \frac{4}{3}

Schritte zur Lösung

\frac{( 2 x + 1 ) ^{2}}{4} + \frac{( 3 y - 2 ) ^{2}}{16} = 1

Rewrite

\frac{( 2 x + 1 ) ^{2}}{4} + \frac{( 3 y - 2 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- \frac{1}{2}, \frac{2}{3}), a = 1, b = \frac{4}{3}

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- \frac{1}{2}, \frac{2}{3}), b = \frac{4}{3}, a = 1

25 x^{2} + 49 y^{2} = 1225

x^{2} - 4 x + 3 y^{2} = - 2

\frac{x ^{2}}{9} + \frac{y ^{2}}{4} - 1 > 0

\frac{x ^{2}}{20} + \frac{y ^{2}}{2} = 1

6 = 4 x^{2} + y^{2} + 1