(a) 证明直角坐标方程为

的曲线，其极坐标方程为 $r^2=36\cos2\theta$。

曲线 $C$ 的极坐标方程为 $r^2=36\cos2\theta$，其中 $-\frac14\pi\le\theta\le\frac14\pi$。

(b) 画出 $C$ 的草图，并写出 $C$ 上的点到极点的最大距离。

(c) 求 $C$ 所围区域的面积。

(d) 求 $C$ 上的点到极轴的最大距离，答案用精确形式表示。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

代入极坐标并用余弦倍角式；除r平方后另外保留极点。

Substituting $x=r\cos\theta$, $y=r\sin\theta$ gives

For $r\ne0$, divide by $r^2$ to obtain $\boxed{r^2=36\cos2\theta}$. The pole is also represented by this polar equation, for example at $\theta=\pi/4$, so it is retained.

角区间只对应右侧一个环，关于极轴对称。极径在θ=0最大，不画左侧另一个环。

The radius is zero at $\theta=\pm\pi/4$ and positive between them. Sketch a single loop to the right of the pole, symmetric about the initial line, passing through $(6,0)$ and returning to the pole. Its two tangents at the pole are the lines $y=\pm x$.

Since $r^2\leqslant36$, the greatest distance from the pole is $\boxed{6}$, at $\theta=0$.

给定角区间完整扫过右环一次，直接积分半个极径平方。

利用上下对称，只在上半环找最大纵坐标。对y求导后，分子可化成余弦三倍角，内部唯一最大点再与端点比较。

By symmetry, the maximum perpendicular distance from the initial line equals the maximum $y$ for $0\leqslant\theta\leqslant\pi/4$, where

For $0<\theta<\pi/4$,

Thus the only interior stationary point is $\theta=\pi/6$. The endpoint values are both $y=0$, whereas at this point

Hence the maximum distance is

It occurs at $\theta=\pm\pi/6$, on the upper and lower halves respectively.