4x^2+2y^2-8x+12y+6=0

解答

4 x^{2} + 2 y^{2} - 8 x + 12 y + 6 = 0

(h, k) = (1, - 3), a = 2, b = 22

求解步骤

4 x^{2} + 2 y^{2} - 8 x + 12 y + 6 = 0

将

4 x^{2} + 2 y^{2} - 8 x + 12 y + 6 = 0

改写为椭圆标准方程

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

因此椭圆的性质为:

(h, k) = (1, - 3), a = 2, b = 22

b > a

因此

b

是长半轴，

a

是短半轴

中心 (h, k) = (1, - 3), 长半轴为 b = 22, 短半轴为 a = 2 的椭圆

\frac{( x + 3 ) ^{2}}{25} + \frac{( y - 8 ) ^{2}}{16} = 1

16 x^{2} + 9 y^{2} - 36 y = 108

3 x^{2} + 12 y^{2} - 6 x - 48 y + 33 = 0

x^{(2)} + 4 y^{(2)} + 2 x - 12 y + 6 = 0

2 x - x^{2} - 4 y^{2} - 16 y - 1 > 0