(24-x^2)+(4-y)^2=(x^2+(y+3)^2)/(23^2)

解答

(24 - x^{2}) + (4 - y)^{2} = \frac{x ^{2} + ( y + 3 ) ^{2}}{2 3 ^{2}}

(h, k) = (0, \frac{2119}{528}), a = \frac{23 12623}{4 17490}, b = \frac{23 12623}{528}

求解步骤

24 - x^{2} + (4 - y)^{2} = \frac{x ^{2} + ( y + 3 ) ^{2}}{2 3 ^{2}}

将

24 - x^{2} + (4 - y)^{2} - \frac{x ^{2} + ( y + 3 ) ^{2}}{2 3 ^{2}} = 0

改写为标准双曲线方程形式

\frac{( x - 0 ) ^{2}}{( \frac{23 12623}{4 17490} ) ^{2}} - \frac{( y - \frac{2119}{528} ) ^{2}}{( \frac{23 12623}{528} ) ^{2}} = 1

因此双曲线的性质为:

(h, k) = (0, \frac{2119}{528}), a = \frac{23 12623}{4 17490}, b = \frac{23 12623}{528}

(y - 1)^{2} - (x - 2)^{2} = 1

4 x^{2} - 9 y^{2} + 8 x - 36 y = 0

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

\frac{x ^{2}}{25} - \frac{y ^{2}}{81} = 1

\frac{( y - 2 ) ^{2}}{9} - \frac{( x + 1 ) ^{2}}{4} = 1