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IAL 2023 May Q10

A Level / Edexcel / P1

IAL 2023 May Paper · Question 10

题目

Problem

Figure 5 shows a sketch of the quadratic curve CC with equation

y=−14(x+2)(x−b)\begin{align*} y=-\frac14(x+2)(x-b) \end{align*}

Figure 5

where bb is a positive constant.

The line l1l_1, also shown in Figure 5,

  • has gradient 12\dfrac12
  • intersects CC on the negative xx-axis and at the point PP

(a) (i) Write down an equation for l1l_1.

(1)

(ii) Find, in terms of bb, the coordinates of PP.

(3)

Given that the line l2l_2 is perpendicular to l1l_1 and intersects CC on the positive xx-axis,

(b) find, in terms of bb, an equation for l2l_2.

(2)

Given also that l2l_2 intersects CC at the point PP,

(c) show that another equation for l2l_2 is

y=−2x+5b2−4.\begin{align*} y=-2x+\frac{5b}{2}-4. \end{align*}
(2)

(d) Hence, or otherwise, find the value of bb.

(2)

解答

(a)(i)

解法一

思路

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曲线的负 xx 轴截距来自 x+2=0x+2=0，所以是 (−2,0)(-2,0)。直线 l1l_1 过这个点，斜率为 12\frac12。

答题过程

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Since l1l_1 passes through (−2,0)(-2,0) and has gradient 12\frac12,

y= 12(x+2)= 12x+1.\begin{align*} y=&\,\frac12(x+2)\\ =&\,\frac12x+1. \end{align*}

(a)(ii)

解法一

思路

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点 PP 是直线 l1l_1 与曲线的另一个交点。联立直线和曲线，除了 x=−2x=-2 这个交点以外，另一个 xx 值就是 PP 的横坐标。

答题过程

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At intersections of l1l_1 and CC,

−14(x+2)(x−b)=12(x+2).\begin{align*} -\frac14(x+2)(x-b)=\frac12(x+2). \end{align*}

One solution is x=−2x=-2. For the other intersection, divide by x+2x+2:

−14(x−b)= 12x−b= −2x= b−2.\begin{align*} -\frac14(x-b)=&\,\frac12\\ x-b=&\,-2\\ x=&\,b-2. \end{align*}

Then

y= 12((b−2)+2)= b2.\begin{align*} y =&\,\frac12((b-2)+2)\\ =&\,\frac b2. \end{align*}

Therefore

P=(b−2,b2).\begin{align*} P=\left(b-2,\frac b2\right). \end{align*}

(b)

解法一

思路

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l1l_1 的斜率是 12\frac12，所以垂直线 l2l_2 的斜率是 −2-2。它与曲线在正 xx 轴相交，即经过 (b,0)(b,0)。

答题过程

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The gradient of l2l_2 is −2-2.

Since l2l_2 passes through (b,0)(b,0),

y−0= −2(x−b)y= −2x+2b.\begin{align*} y-0=&\,-2(x-b)\\ y=&\,-2x+2b. \end{align*}

(c)

解法一

思路

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题目还告诉我们 l2l_2 经过 PP。用 P(b−2,b2)P\left(b-2,\frac b2\right) 和斜率 −2-2 写直线，就会得到另一个形式。

答题过程

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Using gradient −2-2 and

P=(b−2,b2),\begin{align*} P=\left(b-2,\frac b2\right), \end{align*}

the equation of l2l_2 is

y−b2= −2(x−(b−2))y−b2= −2x+2b−4y= −2x+5b2−4.\begin{align*} y-\frac b2=&\,-2(x-(b-2))\\ y-\frac b2=&\,-2x+2b-4\\ y=&\,-2x+\frac{5b}{2}-4. \end{align*}

(d)

解法一

思路

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(b) 和 (c) 给的是同一条直线，所以它们的截距必须相同。比较 −2x+2b-2x+2b 与 −2x+5b2−4-2x+\frac{5b}{2}-4 即可。

答题过程

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Equate the two expressions for l2l_2:

−2x+2b= −2x+5b2−42b= 5b2−44b= 5b−8b= 8.\begin{align*} -2x+2b=&\,-2x+\frac{5b}{2}-4\\ 2b=&\,\frac{5b}{2}-4\\ 4b=&\,5b-8\\ b=&\,8. \end{align*}