4x^2+9y^2+16x-36y+16=0

Lösung

4 x^{2} + 9 y^{2} + 16 x - 36 y + 16 = 0

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

Schritte zur Lösung

4 x^{2} + 9 y^{2} + 16 x - 36 y + 16 = 0

Rewrite

4 x^{2} + 9 y^{2} + 16 x - 36 y + 16 = 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

- 16 - x^{2} - y^{2} \geq 0

\frac{( x - 2 ) ^{2}}{9} + \frac{( y - 1 ) ^{2}}{36} = 1

\frac{( x - 3 ) ^{2}}{16} + \frac{( y + 2 ) ^{2}}{25} = 1

9 x^{2} + 25 y^{2} - 54 x + 100 y - 44 = 0

2 x^{2} - 8 x + y^{2} - 6 y + 7 = 0