1x^2+3y^2+10x+10y-11=0

解答

1 x^{2} + 3 y^{2} + 10 x + 10 y - 11 = 0

(h, k) = (- 5, - \frac{5}{3}), a = \frac{133}{3}, b = \frac{133}{3}

求解步骤

1 \cdot x^{2} + 3 y^{2} + 10 x + 10 y - 11 = 0

将

1 x^{2} + 3 y^{2} + 10 x + 10 y - 11 = 0

改写为椭圆标准方程

\frac{( x - ( - 5 ) ) ^{2}}{( \frac{133}{3} ) ^{2}} + \frac{( y - ( - \frac{5}{3} ) ) ^{2}}{( \frac{133}{3} ) ^{2}} = 1

因此椭圆的性质为:

(h, k) = (- 5, - \frac{5}{3}), a = \frac{133}{3}, b = \frac{133}{3}

a > b

因此

a

是长半轴，

b

是短半轴

中心 (h, k) = (- 5, - \frac{5}{3}), 长半轴为 a = \frac{133}{3}, 短半轴为 b = \frac{133}{3} 的椭圆

16 x^{2} - 160 x + 25 y^{2} + 350 y - 1225 = 0

9 x^{2} + 16 y^{2} - 18 x - 64 y = 71

21 y^{2} + 210 y + 25 x^{2} - 100 x = 0

(\frac{x - 1}{2})^{2} + (y + 3)^{2} = \frac{49}{64}

x^{2} + y^{2} = 2 \cdot \frac{y}{x ^{2} + y ^{2}}