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IAL 2024 Oct Q3

A Level / Edexcel / P4

IAL 2024 Oct Paper · Question 3

题目

Problem

Figure 1 shows a sketch of the curve CC with parametric equations

x=3sin⁡3θy=1+cos⁡2θ−π2≤θ≤π2x=3\sin^3\theta\qquad y=1+\cos 2\theta\qquad -\frac{\pi}{2}\le \theta \le \frac{\pi}{2}

(a) Show that

dydx=kcosec⁡θθ≠0\frac{dy}{dx}=k\operatorname{cosec}\theta\qquad \theta\ne 0

where kk is a constant to be found.

(3)

The point PP lies on CC where θ=π6\theta=\dfrac{\pi}{6}.

(b) Find the equation of the tangent to CC at PP, giving your answer in the form ax+by+c=0ax+by+c=0 where a,ba,b and cc are integers.

(3)

(c) Show that CC has Cartesian equation

8x2=9(2−y)3−q≤x≤q8x^2=9(2-y)^3\qquad -q\le x\le q

where qq is a constant to be found.

(3)
题目中文翻译

图 1 给出了曲线 CC 的草图，其参数方程为

x=3sin⁡3θy=1+cos⁡2θ−π2≤θ≤π2x=3\sin^3\theta\qquad y=1+\cos 2\theta\qquad -\frac{\pi}{2}\le \theta \le \frac{\pi}{2}

(a) 证明

dydx=kcosec⁡θθ≠0\frac{dy}{dx}=k\operatorname{cosec}\theta\qquad \theta\ne 0

其中 kk 是待求常数。

点 PP 在曲线 CC 上，且 θ=π6\theta=\dfrac{\pi}{6}。

(b) 求曲线 CC 在点 PP 处的切线方程，答案写成 ax+by+c=0ax+by+c=0 的形式，其中 a,b,ca,b,c 为整数。

(c) 证明曲线 CC 的直角坐标方程为

8x2=9(2−y)3−q≤x≤q8x^2=9(2-y)^3\qquad -q\le x\le q

其中 qq 是待求常数。

解答

(a)

We are given

x=3sin⁡3θx=3\sin^3\theta

and

y=1+cos⁡2θy=1+\cos2\theta

Differentiate with respect to θ\theta:

dxdθ=9sin⁡2θcos⁡θ\frac{dx}{d\theta}=9\sin^2\theta\cos\theta

and

dydθ=−2sin⁡2θ\frac{dy}{d\theta}=-2\sin2\theta

Using

sin⁡2θ=2sin⁡θcos⁡θ\sin2\theta=2\sin\theta\cos\theta

we have

dydθ=−4sin⁡θcos⁡θ\frac{dy}{d\theta}=-4\sin\theta\cos\theta

Therefore

dydx=dydθdxdθ=−4sin⁡θcos⁡θ9sin⁡2θcos⁡θ\frac{dy}{dx} =\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} =\frac{-4\sin\theta\cos\theta}{9\sin^2\theta\cos\theta}

So

dydx=−49sin⁡θ=−49cosec⁡θ\frac{dy}{dx} =-\frac{4}{9\sin\theta} =-\frac49\operatorname{cosec}\theta

Hence

k=−49\boxed{k=-\frac49}

(b)

At PP,

θ=π6\theta=\frac{\pi}{6}

First find the coordinates of PP.

x=3sin⁡3π6=3(12)3=38x=3\sin^3\frac{\pi}{6} =3\left(\frac12\right)^3 =\frac38

Also,

y=1+cos⁡π3=1+12=32y=1+\cos\frac{\pi}{3} =1+\frac12 =\frac32

Now find the gradient:

dydx=−49cosec⁡π6=−49⋅2=−89\frac{dy}{dx} =-\frac49\operatorname{cosec}\frac{\pi}{6} =-\frac49\cdot2 =-\frac89

So the tangent at PP is

y−32=−89(x−38)y-\frac32=-\frac89\left(x-\frac38\right)

Multiply by 1818:

18y−27=−16x+618y-27=-16x+6

Therefore

16x+18y−33=0\boxed{16x+18y-33=0}

(c)

From

x=3sin⁡3θx=3\sin^3\theta

we get

x2=9sin⁡6θx^2=9\sin^6\theta

So

8x2=72sin⁡6θ8x^2=72\sin^6\theta

Now

y=1+cos⁡2θy=1+\cos2\theta

Using

cos⁡2θ=1−2sin⁡2θ\cos2\theta=1-2\sin^2\theta

we get

y=1+1−2sin⁡2θy=1+1-2\sin^2\theta

Thus

2−y=2sin⁡2θ2-y=2\sin^2\theta

Therefore

9(2−y)3=9(2sin⁡2θ)3=72sin⁡6θ9(2-y)^3 =9(2\sin^2\theta)^3 =72\sin^6\theta

Hence

8x2=9(2−y)38x^2=9(2-y)^3

Also, since

−π2≤θ≤π2-\frac{\pi}{2}\le \theta\le \frac{\pi}{2}

we have

−1≤sin⁡θ≤1-1\le \sin\theta\le 1

So

−3≤3sin⁡3θ≤3-3\le 3\sin^3\theta\le 3

and hence

−3≤x≤3-3\le x\le 3

Therefore

q=3\boxed{q=3}