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IAL 2025 Jan Q9

A Level / Edexcel / P4

IAL 2025 Jan Paper · Question 9

题目

Problem

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

Figure 2 shows a sketch of the curve CC with parametric equations

x=2cos⁡2ty=sin⁡3t−π2<t<π2x=2\cos 2t\qquad y=\sin^3 t\qquad -\frac{\pi}{2}<t<\frac{\pi}{2}

where tt is a parameter.

The point PP lies on CC where t=π6t=\dfrac{\pi}{6}.

The line ll, shown in Figure 2, is the tangent to CC at PP.

(a) Use parametric differentiation to show that

(i) dydx=ksin⁡t\dfrac{dy}{dx}=k\sin t where kk is a constant to be found

(ii) an equation for ll is 3x+16y−5=03x+16y-5=0

(6)

The line ll intersects the curve CC again at the point QQ.

(b) Using algebra and showing detailed reasoning, find the exact coordinates of QQ.

(6)
题目中文翻译

本题必须写出全部解题步骤。

不接受完全依赖计算器技术的解法。

图 2 给出了曲线 CC 的草图，其参数方程为

x=2cos⁡2ty=sin⁡3t−π2<t<π2x=2\cos 2t\qquad y=\sin^3 t\qquad -\frac{\pi}{2}<t<\frac{\pi}{2}

其中 tt 为参数。

点 PP 在曲线 CC 上，且 t=π6t=\dfrac{\pi}{6}。

图 2 中的直线 ll 是曲线 CC 在点 PP 处的切线。

(a) 用参数求导证明：

(i) dydx=ksin⁡t\dfrac{dy}{dx}=k\sin t，其中 kk 是待求常数；

(ii) 直线 ll 的一个方程为 3x+16y−5=03x+16y-5=0。

直线 ll 再次与曲线 CC 交于点 QQ。

(b) 用代数方法并写出详细推理，求点 QQ 的精确坐标。

解答

(a)(i)

We are given

x=2cos⁡2t,y=sin⁡3tx=2\cos 2t,\qquad y=\sin^3t

Differentiate with respect to tt:

dxdt=−4sin⁡2t\frac{dx}{dt}=-4\sin2t

and

dydt=3sin⁡2tcos⁡t\frac{dy}{dt}=3\sin^2t\cos t

Therefore

dydx=dydtdxdt=3sin⁡2tcos⁡t−4sin⁡2t\frac{dy}{dx} =\frac{\frac{dy}{dt}}{\frac{dx}{dt}} =\frac{3\sin^2t\cos t}{-4\sin2t}

Using

sin⁡2t=2sin⁡tcos⁡t\sin2t=2\sin t\cos t

we get

dydx=3sin⁡2tcos⁡t−8sin⁡tcos⁡t\frac{dy}{dx} =\frac{3\sin^2t\cos t}{-8\sin t\cos t}

Hence

dydx=−38sin⁡t\frac{dy}{dx}=-\frac38\sin t

So

k=−38\boxed{k=-\frac38}

(a)(ii)

At PP,

t=π6t=\frac{\pi}{6}

So

x=2cos⁡π3=1x=2\cos\frac{\pi}{3}=1

and

y=sin⁡3π6=(12)3=18y=\sin^3\frac{\pi}{6} =\left(\frac12\right)^3 =\frac18

Therefore

P=(1,18)P=\left(1,\frac18\right)

The gradient of the tangent at PP is

dydx=−38sin⁡π6=−38⋅12=−316\frac{dy}{dx} =-\frac38\sin\frac{\pi}{6} =-\frac38\cdot\frac12 =-\frac{3}{16}

Using the point-gradient form,

y−18=−316(x−1)y-\frac18=-\frac{3}{16}(x-1)

Multiply by 1616:

16y−2=−3x+316y-2=-3x+3

Hence

3x+16y−5=0\boxed{3x+16y-5=0}

as required.

(b)

At an intersection of the line ll and the curve CC,

3x+16y−5=03x+16y-5=0

Substitute

x=2cos⁡2t,y=sin⁡3tx=2\cos2t,\qquad y=\sin^3t

to get

3(2cos⁡2t)+16sin⁡3t−5=03(2\cos2t)+16\sin^3t-5=0

So

6cos⁡2t+16sin⁡3t−5=06\cos2t+16\sin^3t-5=0

Use

cos⁡2t=1−2sin⁡2t\cos2t=1-2\sin^2t

Then

6(1−2sin⁡2t)+16sin⁡3t−5=06(1-2\sin^2t)+16\sin^3t-5=0

So

16sin⁡3t−12sin⁡2t+1=016\sin^3t-12\sin^2t+1=0

Let

u=sin⁡tu=\sin t

Then

16u3−12u2+1=016u^3-12u^2+1=0

Since PP corresponds to u=12u=\dfrac12, factor out 2u−12u-1:

16u3−12u2+1=(2u−1)(8u2−2u−1)16u^3-12u^2+1=(2u-1)(8u^2-2u-1)

Now

8u2−2u−1=(4u+1)(2u−1)8u^2-2u-1=(4u+1)(2u-1)

Therefore

16u3−12u2+1=(2u−1)2(4u+1)16u^3-12u^2+1=(2u-1)^2(4u+1)

For the second intersection QQ,

4u+1=04u+1=0

so

sin⁡t=−14\sin t=-\frac14

Now

y=sin⁡3t=(−14)3=−164y=\sin^3t=\left(-\frac14\right)^3=-\frac{1}{64}

Also,

cos⁡2t=1−2sin⁡2t=1−2(116)=78\cos2t=1-2\sin^2t =1-2\left(\frac{1}{16}\right) =\frac78

Thus

x=2cos⁡2t=2⋅78=74x=2\cos2t=2\cdot\frac78=\frac74

Therefore

Q=(74,−164)\boxed{Q=\left(\frac74,-\frac{1}{64}\right)}