7(x-2004)^2=23-y^2

解答

7 (x - 2004)^{2} = 23 - y^{2}

(h, k) = (2004, 0), a = \frac{23}{7}, b = 23

求解步骤

7 (x - 2004)^{2} = 23 - y^{2}

将

7 (x - 2004)^{2} - 23 + y^{2} = 0

改写为椭圆标准方程

\frac{( x - 2004 ) ^{2}}{( \frac{23}{7} ) ^{2}} + \frac{( y - 0 ) ^{2}}{( 23 ) ^{2}} = 1

因此椭圆的性质为:

(h, k) = (2004, 0), a = \frac{23}{7}, b = 23

b > a

因此

b

是长半轴，

a

是短半轴

中心 (h, k) = (2004, 0), 长半轴为 b = 23, 短半轴为 a = \frac{23}{7} 的椭圆

\frac{( x - 6 ) ^{2}}{25} + \frac{( y - 7 ) ^{2}}{9} = 1

8 x^{2} + y^{2} - 32 x - 16 y = - 16

\frac{x ^{2}}{16} + \frac{y ^{2}}{48} = 1

\frac{( x - 2 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{5} = 1

16 (x + 1)^{2} + y^{2} = 16