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IAL 2024 Jan Q9

A Level / Edexcel / P4

IAL 2024 Jan Paper · Question 9

题目

Problem

Figure 4 shows a sketch of the curve CC with parametric equations

x=sec⁡ty=3tan⁡(t+π3)π6<t<π2x=\sec t\qquad y=\sqrt{3}\tan\left(t+\frac{\pi}{3}\right)\qquad \frac{\pi}{6}<t<\frac{\pi}{2}

(a) Find dydx\dfrac{dy}{dx} in terms of tt.

(3)

(b) Find an equation for the tangent to CC at the point where t=π3t=\dfrac{\pi}{3}.

Give your answer in the form y=mx+cy=mx+c, where mm and cc are constants.

(4)

(c) Show that all points on CC satisfy the equation

y=Ax2+B3x2−34−3x2y=\frac{Ax^2+B\sqrt{3x^2-3}}{4-3x^2}

where AA and BB are constants to be found.

(5)
题目中文翻译

图 4 给出了曲线 CC 的草图，其参数方程为

x=sec⁡ty=3tan⁡(t+π3)π6<t<π2x=\sec t\qquad y=\sqrt{3}\tan\left(t+\frac{\pi}{3}\right)\qquad \frac{\pi}{6}<t<\frac{\pi}{2}

(a) 用 tt 表示 dydx\dfrac{dy}{dx}。

(b) 求当 t=π3t=\dfrac{\pi}{3} 时曲线 CC 的切线方程。

答案写成 y=mx+cy=mx+c 的形式，其中 m,cm,c 为常数。

(c) 证明曲线 CC 上所有点都满足方程

y=Ax2+B3x2−34−3x2y=\frac{Ax^2+B\sqrt{3x^2-3}}{4-3x^2}

其中 A,BA,B 是待求常数。

解答

(a)

解法一

思路

展开

分别对两个参数方程关于 tt 求导，再使用

dydx=dydt÷dxdt.\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{\mathrm{d}y}{\mathrm{d}t} \div \frac{\mathrm{d}x}{\mathrm{d}t}.

答题过程

展开

Differentiating the parametric equations with respect to tt gives

dxdt=sec⁡ttan⁡t\frac{\mathrm{d}x}{\mathrm{d}t} =\sec t\tan t

and

dydt=3sec⁡2(t+π3).\frac{\mathrm{d}y}{\mathrm{d}t} =\sqrt{3}\sec^2\bigg(t+\frac{\pi}{3}\bigg).

Therefore,

dydx= dydt÷dxdt= 3sec⁡2(t+π3)sec⁡ttan⁡t.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\frac{\mathrm{d}y}{\mathrm{d}t} \div\frac{\mathrm{d}x}{\mathrm{d}t}\\ =&\,\boxed{ \frac{\sqrt{3}\sec^2\big(t+\frac{\pi}{3}\big)} {\sec t\tan t} }. \end{align*}

(b)

解法一

思路

展开

把 t=π3t=\frac{\pi}{3} 分别代入 xx、yy 和上一问的导数，求出切点及切线斜率，再使用点斜式写出切线方程。

答题过程

展开

When t=π3t=\frac{\pi}{3},

x=sec⁡π3=2x=\sec\frac{\pi}{3}=2

and

y=3tan⁡2π3=−3.y=\sqrt{3}\tan\frac{2\pi}{3}=-3.

The gradient of the tangent is

dydx= 3sec⁡22π3sec⁡π3tan⁡π3= 4323= 2.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\frac{ \sqrt{3}\sec^2\frac{2\pi}{3} }{ \sec\frac{\pi}{3}\tan\frac{\pi}{3} }\\ =&\,\frac{4\sqrt{3}}{2\sqrt{3}}\\ =&\,2. \end{align*}

Using the point (2,−3)(2,-3),

y+3= 2(x−2),y= 2x−7.\begin{align*} y+3=&\,2(x-2),\\ y=&\,\boxed{2x-7}. \end{align*}

(c)

解法一

思路

展开

先用正切和角公式展开 yy。由 x=sec⁡tx=\sec t 可得 tan⁡2t=x2−1\tan^2t=x^2-1；又因为题设范围内 t>0t>0，所以取 tan⁡t=x2−1\tan t=\sqrt{x^2-1}。代入后对分母有理化，即可整理成题目指定的形式。

答题过程

展开

Using the tangent addition formula,

y= 3tan⁡t+tan⁡π31−tan⁡ttan⁡π3= 3tan⁡t+31−3tan⁡t.\begin{align*} y =&\,\sqrt{3} \frac{\tan t+\tan\frac{\pi}{3}} {1-\tan t\tan\frac{\pi}{3}}\\ =&\,\sqrt{3} \frac{\tan t+\sqrt{3}} {1-\sqrt{3}\tan t}. \end{align*}

Also,

x2=sec⁡2t=1+tan⁡2t.x^2=\sec^2t=1+\tan^2t.

Since π6<t<π2\frac{\pi}{6}<t<\frac{\pi}{2}, tan⁡t>0\tan t>0, and hence

tan⁡t=x2−1.\tan t=\sqrt{x^2-1}.

Let u=x2−1u=\sqrt{x^2-1}. Then

y=3u+31−3u.y=\frac{\sqrt{3}u+3}{1-\sqrt{3}u}.

Multiplying the numerator and denominator by 1+3u1+\sqrt{3}u,

y= (3u+3)(1+3u)(1−3u)(1+3u)= 3u2+3+43u1−3u2.\begin{align*} y =&\,\frac{ (\sqrt{3}u+3)(1+\sqrt{3}u) }{ (1-\sqrt{3}u)(1+\sqrt{3}u) }\\ =&\,\frac{3u^2+3+4\sqrt{3}u}{1-3u^2}. \end{align*}

Substituting u2=x2−1u^2=x^2-1 and u=x2−1u=\sqrt{x^2-1} gives

y= 3(x2−1)+3+43x2−11−3(x2−1)= 3x2+43x2−34−3x2.\begin{align*} y =&\,\frac{ 3(x^2-1)+3 +4\sqrt{3}\sqrt{x^2-1} }{ 1-3(x^2-1) }\\ =&\,\frac{ 3x^2+4\sqrt{3x^2-3} }{ 4-3x^2 }. \end{align*}

This is the required form, with

A=3,B=4.\boxed{A=3,\qquad B=4}.