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CIE 9231 2023 June Paper 13 Q5

A Level / CIE / FP1

CIE 9231 2023 June Paper 13 Paper · Question 5

题目

Problem

(a) Show that the curve with Cartesian equation

x2−y2=a,x^2-y^2=a,

where aa is a positive constant, has polar equation r2=asec⁡2θr^2=a\sec2\theta.

(3)

The curve CC has polar equation r2=asec⁡2θr^2=a\sec2\theta, where aa is a positive constant, for 0≤θ<14π0\le\theta<\frac14\pi.

(b) Sketch CC and state the minimum distance of CC from the pole.

(3)

(c) Find, in terms of aa, the exact value of the area of the region enclosed by CC, the initial line, and the half-line θ=112π\theta=\frac1{12}\pi. [You may use any result from the list of formulae (MF19) without proof.]

(4)
题目中文翻译

(a) 证明，笛卡尔方程为

x2−y2=a,x^2-y^2=a,

的曲线（其中 aa 为正常数），其极坐标方程为 r2=asec⁡2θr^2=a\sec2\theta。

(3)

曲线 CC 的极坐标方程为 r2=asec⁡2θr^2=a\sec2\theta，其中 aa 为正常数，且 0≤θ<14π0\le\theta<\frac14\pi。

(b) 画出 CC 的示意图，并写出 CC 到极点的最短距离。

(3)

(c) 求由 CC、极轴及射线 θ=112π\theta=\frac1{12}\pi 围成的区域面积的精确值，用 aa 表示。[你可以使用公式表（MF19）中的任何结果，无须证明。]

(4)
原卷通用作答要求

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。

解答

(a)

解法一

思路

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把平方坐标代入极坐标，再用余弦倍角公式。a为正保证曲线上的r和cos2θ均不为零。

答题过程

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Substituting x=rcos⁡θx=r\cos\theta, y=rsin⁡θy=r\sin\theta gives

r2(cos⁡2θ−sin⁡2θ)= a,r2cos⁡2θ= a.\begin{align*} r^2(\cos^2\theta-\sin^2\theta)=&\,a,\\[4mm] r^2\cos2\theta=&\,a. \end{align*}

Since a>0a>0, cos⁡2θ≠0\cos2\theta\ne0 on the curve. Therefore

r2=asec⁡2θ.\boxed{r^2=a\sec2\theta}.

(b)

解法一

思路

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这是双曲线右支的第一象限部分，起点到极点的距离最小。越靠近四十五度射线，半径越大，曲线渐近于y=x。

答题过程

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On 0⩽θ<π/40\leqslant\theta<\pi/4, cos⁡2θ\cos2\theta decreases from 11 to 0+0^+, so rr increases from a\sqrt a to +∞+\infty.

Sketch the first-quadrant part of x2−y2=ax^2-y^2=a, starting at (a,0)(\sqrt a,0) with vertical tangent and bending to the right. It approaches y=xy=x from below as θ→π/4−\theta\to\pi/4^-.

The minimum distance from the pole is a\boxed{\sqrt a}.

(c)

解法一

思路

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代入极坐标面积，用正割积分公式；角度倍数2使积分再多一个二分之一。

答题过程

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Using the formula for the integral of sec⁡u\sec u,

A= a2∫0π/12sec⁡2θ dθ= a4[ln⁡∣sec⁡2θ+tan⁡2θ∣]0π/12= a4ln⁡(23+13)= a4ln⁡3= a8ln⁡3.\begin{align*} A=&\,\frac a2\int_0^{\pi/12}\sec2\theta\,\mathrm{d}\theta\\[4mm] =&\,\frac a4\big[\ln|\sec2\theta+\tan2\theta|\big]_0^{\pi/12}\\[4mm] =&\,\frac a4\ln\bigg(\frac2{\sqrt3}+\frac1{\sqrt3}\bigg)\\[4mm] =&\,\frac a4\ln\sqrt3\\[4mm] =&\,\boxed{\frac a8\ln3}. \end{align*}