3x^2+4y^2+28x-16y+48=0

Lösung

3 x^{2} + 4 y^{2} + 28 x - 16 y + 48 = 0

(h, k) = (- \frac{14}{3}, 2), a = \frac{10}{3}, b = \frac{5}{3}

Schritte zur Lösung

3 x^{2} + 4 y^{2} + 28 x - 16 y + 48 = 0

Rewrite

3 x^{2} + 4 y^{2} + 28 x - 16 y + 48 = 0

in the form of the standard ellipse equation

\frac{( x - ( - \frac{14}{3} ) ) ^{2}}{( \frac{10}{3} ) ^{2}} + \frac{( y - 2 ) ^{2}}{( \frac{5}{3} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- \frac{14}{3}, 2), a = \frac{10}{3}, b = \frac{5}{3}

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- \frac{14}{3}, 2), a = \frac{10}{3}, b = \frac{5}{3}

5 x^{2} + 9 y^{2} + 30 x = 0

4 x^{2} + 7 y^{2} = 28

3 x^{2} + 2 y^{2} - 6 x - 12 y + 15 = 0

9 y^{2} + x^{2} = 9

4 x^{2} + 7 y^{2} - 140 = 0