Sketch the curve CC with equation x2−(y+1)2=1x^2 - (y + 1)^2 = 1, indicating clearly the coordinates of any points of intersection with the axes and the equations of any asymptotes. The curve CC undergoes a sequence of three transformations as follows: I: A scaling parallel to xx-axis with a scale factor of 4,
II: A scaling parallel to yy-axis with a scale factor of 12\frac{1}{2},
III: A translation of 2 units in the positive yy-direction.
Find the equation of the resulting curve.

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Worked solution

1. 求曲线 $C$ 与坐标轴的交点坐标以及渐近线方程

曲线 CC 的方程为 x2−(y+1)2=1x^2 - (y + 1)^2 = 1。

  1. 与 xx 轴的交点:令 y=0y = 0,得 x2−(0+1)2=1⇒x2=2⇒x=±2x^2 - (0 + 1)^2 = 1 \Rightarrow x^2 = 2 \Rightarrow x = \pm\sqrt{2}。
    故与 xx 轴的交点坐标为 (2,0)(\sqrt{2}, 0) 和 (−2,0)(-\sqrt{2}, 0)。
  2. 与 yy 轴的交点:令 x=0x = 0,得 −(y+1)2=1⇒(y+1)2=−1-(y + 1)^2 = 1 \Rightarrow (y + 1)^2 = -1,无实数解,故曲线与 yy 轴没有交点。
  3. 渐近线方程:令 x2−(y+1)2=0x^2 - (y + 1)^2 = 0,得 y+1=±xy + 1 = \pm x,即渐近线方程为 y=x−1y = x - 1 和 y=−x−1y = -x - 1。

2. 绘制曲线 $C$ 的草图

曲线 CC 为双曲线,中心在 (0,−1)(0, -1),顶点坐标为 (1,−1)(1, -1) 和 (−1,−1)(-1, -1)。
画图时需清晰标出:

  • 顶点 (1,−1)(1, -1) 和 (−1,−1)(-1, -1);
  • 与 xx 轴的交点 (2,0)(\sqrt{2}, 0) 和 (−2,0)(-\sqrt{2}, 0);
  • 两条渐近线 y=x−1y = x - 1 和 y=−x−1y = -x - 1。

3. 对曲线 $C$ 依次施加三种变换并求出最终方程

原曲线 CC 方程为 x2−(y+1)2=1x^2 - (y + 1)^2 = 1。

  • 变换 I(沿 xx 轴方向伸缩,伸缩比例为 4):用 x4\dfrac{x}{4} 替换 xx,得到方程:
    (x4)2−(y+1)2=1  ⟹  x216−(y+1)2=1\left(\frac{x}{4}\right)^2 - (y + 1)^2 = 1 \implies \frac{x^2}{16} - (y + 1)^2 = 1
  • 变换 II(沿 yy 轴方向伸缩,伸缩比例为 12\dfrac{1}{2}):用 2y2y 替换 yy,得到方程:
    x216−(2y+1)2=1\frac{x^2}{16} - (2y + 1)^2 = 1
  • 变换 III(沿 yy 轴正方向平移 2 个单位):用 y−2y - 2 替换 yy,得到方程:
    x216−[2(y−2)+1]2=1  ⟹  x216−(2y−3)2=1\frac{x^2}{16} - [2(y - 2) + 1]^2 = 1 \implies \frac{x^2}{16} - (2y - 3)^2 = 1
Answer

曲线 CC 的坐标轴交点为 (2,0)(\sqrt{2}, 0) 和 (−2,0)(-\sqrt{2}, 0),渐近线方程为 y=x−1y = x - 1 和 y=−x−1y = -x - 1(无 yy 轴交点);变换后的曲线方程为 x216−(2y−3)2=1\dfrac{x^2}{16} - (2y - 3)^2 = 1。

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