CalcGospel 國際數學圖譜
返回

IAL 2020 Oct Q4

A Level / Edexcel / P4

IAL 2020 Oct Paper · Question 4

题目

Problem

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

x=2t2−6t,y=t3−4t,t∈Rx=2t^2-6t,\qquad y=t^3-4t,\qquad t\in\mathbb{R}

The curve cuts the xx-axis at the origin and at the points AA and BB, as shown in Figure 2.

(a) Find the coordinates of AA and show that BB has coordinates (20,0)(20,0).

(3)

(b) Show that the equation of the tangent to the curve at BB is

7y+4x−80=07y+4x-80=0
(5)

The tangent to the curve at BB cuts the curve again at the point PP.

(c) Find, using algebra, the xx coordinate of PP.

(4)
题目中文翻译

图 2 给出了曲线的参数方程草图：

x=2t2−6t,y=t3−4t,t∈Rx=2t^2-6t,\qquad y=t^3-4t,\qquad t\in\mathbb{R}

该曲线与 xx 轴相交于原点以及点 AA 和 BB，如图 2 所示。

(a) 求点 AA 的坐标，并证明点 BB 的坐标为 (20,0)(20,0)。

(b) 证明该曲线在点 BB 处的切线方程为

7y+4x−80=07y+4x-80=0

曲线在点 BB 处的切线再次与曲线相交于点 PP。

(c) 用代数方法求点 PP 的 xx 坐标。

解答

(a)

解法一

思路

展开

曲线与 xx 轴相交时 y=0y=0。先解参数方程 t3−4t=0t^3-4t=0，其中 t=0t=0 对应原点；把另外两个参数值代入 x=2t2−6tx=2t^2-6t，便可辨认点 AA 和 BB。

答题过程

展开

At an intersection with the xx-axis,

t3−4t= 0,t(t2−4)= 0.\begin{align*} t^3-4t =&\,0,\\ t(t^2-4) =&\,0. \end{align*}

Thus t=0t=0, t=2t=2 or t=−2t=-2. The value t=0t=0 gives the origin.

When t=2t=2,

x=2(2)2−6(2)=−4,x=2(2)^2-6(2)=-4,

so

A=(−4,0).\boxed{A=(-4,0)}.

When t=−2t=-2,

x= 2(−2)2−6(−2)= 8+12= 20,\begin{align*} x =&\,2(-2)^2-6(-2)\\ =&\,8+12\\ =&\,20, \end{align*}

and y=0y=0. Hence,

B=(20,0),\boxed{B=(20,0)},

as required.

(b)

解法一

思路

展开

点 BB 对应 t=−2t=-2。分别对两个参数方程关于 tt 求导，再使用参数曲线的导数公式求切线斜率，最后用点斜式整理成题目给出的方程。

答题过程

展开

Differentiating the parametric equations,

dxdt= 4t−6,dydt= 3t2−4.\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}t} =&\,4t-6,\\ \frac{\mathrm{d}y}{\mathrm{d}t} =&\,3t^2-4. \end{align*}

Therefore,

dydx=3t2−44t−6.\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{3t^2-4}{4t-6}.

At BB, t=−2t=-2, so the gradient is

m= 3(−2)2−44(−2)−6= 8−14= −47.\begin{align*} m =&\,\frac{3(-2)^2-4}{4(-2)-6}\\ =&\,\frac{8}{-14}\\ =&\,-\frac47. \end{align*}

Using the point B=(20,0)B=(20,0), the tangent is

y=−47(x−20).y=-\frac47(x-20).

Hence,

7y+4x−80=0,\boxed{7y+4x-80=0},

as required.

(c)

解法一

思路

展开

把参数方程中的 xx、yy 代入切线方程，得到关于 tt 的三次方程。由于切线在 BB 处与曲线相切，t=−2t=-2 是重根；因式分解后取代表再次相交点 PP 的另一个参数值，再代回 x(t)x(t)。

答题过程

展开

At an intersection of the curve and the tangent,

7(t3−4t)+4(2t2−6t)−80=0.7(t^3-4t)+4(2t^2-6t)-80=0.

Therefore,

7t3+8t2−52t−80= 0,(t+2)2(7t−20)= 0.\begin{align*} 7t^3+8t^2-52t-80 =&\,0,\\ (t+2)^2(7t-20) =&\,0. \end{align*}

The repeated root t=−2t=-2 corresponds to the point of tangency BB. The other intersection PP therefore corresponds to

t=207.t=\frac{20}{7}.

Hence,

xP= 2(207)2− 6(207)= 80049− 84049= −4049.\begin{align*} x_P =&\,2\bigg(\frac{20}{7}\bigg)^2\\ -&\,6\bigg(\frac{20}{7}\bigg)\\ =&\,\frac{800}{49}\\ -&\,\frac{840}{49}\\ =&\,\boxed{-\frac{40}{49}}. \end{align*}