16x^2+25y^2-32x-100y-284=0

解答

16 x^{2} + 25 y^{2} - 32 x - 100 y - 284 = 0

(h, k) = (1, 2), a = 5, b = 4

求解步骤

16 x^{2} + 25 y^{2} - 32 x - 100 y - 284 = 0

将

16 x^{2} + 25 y^{2} - 32 x - 100 y - 284 = 0

改写为椭圆标准方程

\frac{( x - 1 ) ^{2}}{5 ^{2}} + \frac{( y - 2 ) ^{2}}{4 ^{2}} = 1

因此椭圆的性质为:

(h, k) = (1, 2), a = 5, b = 4

a > b

因此

a

是长半轴，

b

是短半轴

中心 (h, k) = (1, 2), 长半轴为 a = 5, 短半轴为 b = 4 的椭圆

9 x^{2} - 54 x + 4 y^{2} - 8 y - 59 = 0

\frac{( x + 5 ) ^{2}}{20} + \frac{( y - 6 ) ^{2}}{11} = 1

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

x^{2} + y^{2} + y^{2} = \frac{9}{4}

(x - 0.017)^{2} + (\frac{y ^{2}}{( 0.999 ) ^{2}}) = 1