曲线 CC 的参数方程为
x=5asec⁡θ,y=3atan⁡θ,where −π2<θ<π2x = 5a\sec\theta, \quad y = 3a\tan\theta, \quad \text{where } -\frac{\pi}{2} < \theta < \frac{\pi}{2}
其中 aa 是一个正数常数。

  1. (i)

    求 CC 的直角坐标方程。

  2. (ii)

    草拟 CC 的图形,并给出与坐标轴的任何交点坐标以及任何渐近线的方程。

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完整解法

6(i)

1. Express the trigonometric functions in terms of $x$ and $y$.

x=5asec⁡θ  ⟹  sec⁡θ=x5ax = 5a\sec\theta \implies \sec\theta = \frac{x}{5a}
y=3atan⁡θ  ⟹  tan⁡θ=y3ay = 3a\tan\theta \implies \tan\theta = \frac{y}{3a}

2. Eliminate $\theta$ using the Pythagorean identity for $\sec$ and $\tan$.

We know the identity sec⁡2θ−tan⁡2θ=1\sec^2\theta - \tan^2\theta = 1.
Substituting the expressions:
(x5a)2−(y3a)2=1\left(\frac{x}{5a}\right)^2 - \left(\frac{y}{3a}\right)^2 = 1
x225a2−y29a2=1\frac{x^2}{25a^2} - \frac{y^2}{9a^2} = 1

3. Determine the domain restriction of the curve based on the parameter's given interval.

For −π2<θ<π2-\frac{\pi}{2} < \theta < \frac{\pi}{2}, we have sec⁡θ≥1\sec\theta \ge 1.
Since aa is a positive constant, this implies x≥5ax \ge 5a.
(The range of tan⁡θ\tan\theta is R\mathbb{R}, so yy can take any real value).

答案

x225a2−y29a2=1,x≥5a\frac{x^2}{25a^2} - \frac{y^2}{9a^2} = 1, \quad x \ge 5a

6(ii)

1. Find the cartesian equation (from part i) and determine the valid domain of $x$ to identify the correct portion of the curve.

x=5asec⁡θ  ⟹  sec⁡θ=x5ax = 5a\sec\theta \implies \sec\theta = \frac{x}{5a}
y=3atan⁡θ  ⟹  tan⁡θ=y3ay = 3a\tan\theta \implies \tan\theta = \frac{y}{3a}
Using the identity sec⁡2θ−tan⁡2θ=1\sec^2\theta - \tan^2\theta = 1:
x225a2−y29a2=1\frac{x^2}{25a^2} - \frac{y^2}{9a^2} = 1
Given the parameter domain −π2<θ<π2-\frac{\pi}{2} < \theta < \frac{\pi}{2}, we have sec⁡θ≥1\sec\theta \ge 1.
Since a>0a > 0, this restricts xx to x≥5ax \ge 5a.
Thus, CC represents only the right branch of the hyperbola.

2. Find the coordinates of the intersection with the coordinate axes.

For yy-intercepts, substitute x=0x = 0. Since x≥5ax \ge 5a and a>0a > 0, there are no intersections with the yy-axis.
For xx-intercepts, substitute y=0y = 0:
x225a2−0=1  ⟹  x2=25a2\frac{x^2}{25a^2} - 0 = 1 \implies x^2 = 25a^2
Since x≥5ax \ge 5a, we take the positive root: x=5ax = 5a.
The intersection with the axes is (5a,0)(5a, 0).

3. Determine the equations of the asymptotes.

For a standard hyperbola x2A2−y2B2=1\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1, the asymptotes are given by y=±BAxy = \pm \frac{B}{A}x.
Here, A=5aA = 5a and B=3aB = 3a.
The asymptotes are y=±3a5axy = \pm \frac{3a}{5a}x, which simplifies to:
y=35xandy=−35xy = \frac{3}{5}x \quad \text{and} \quad y = -\frac{3}{5}x

4. Sketch the curve $C$ incorporating all the established features.

Sketch requirements:

  • Draw dashed lines for the asymptotes y=35xy = \frac{3}{5}x and y=−35xy = -\frac{3}{5}x passing through the origin.
  • Mark the xx-intercept at (5a,0)(5a, 0).
  • Draw a smooth continuous curve representing the right branch of the hyperbola, opening towards the right.
  • The curve must originate from the vertex (5a,0)(5a, 0) and approach the asymptotes as x→∞x \to \infty.
答案

Intersection with the axes: (5a,0)(5a, 0)
Asymptotes: y=35xy = \frac{3}{5}x and y=−35xy = -\frac{3}{5}x

(A graph sketch showing the right branch of the hyperbola starting at (5a,0)(5a, 0) and approaching the dotted asymptotes y=±35xy = \pm \frac{3}{5}x)

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