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FP1 2021 June Q6

A Level / Edexcel / FP1

IAL 2021 June Paper · Question 6

题目

Problem

6. The parabola CC has Cartesian equation y2=8xy^2 = 8x

The point P(2p2,4p)P(2p^2, 4p) and the point Q(2q2,4q)Q(2q^2, 4q), where p,q≠0p, q \neq 0, p≠qp \neq q, are points on CC.

(a) Show that an equation of the normal to CC at PP is

y+px=2p3+4py + px = 2p^3 + 4p
(5)

(b) Write down an equation of the normal to CC at QQ

(1)

The normal to CC at PP and the normal to CC at QQ meet at the point NN

(c) Show that NN has coordinates

 (2(p2+pq+q2+2), −2pq(p+q))\begin{align*} &\,\big(2(p^2+pq+q^2+2),\\ &\,-2pq(p+q)\big) \end{align*}
(5)

The line ONON, where OO is the origin, is perpendicular to the line PQPQ

(d) Find the value of (p+q)2−3pq(p + q)^2 - 3pq

(5)
题目中文翻译
  1. 抛物线 CC 的直角坐标方程为 y2=8xy^2 = 8x

点 P(2p2,4p)P(2p^2, 4p) 和点 Q(2q2,4q)Q(2q^2, 4q) 在 CC 上，其中 p,q≠0p, q \neq 0，p≠qp \neq q。

(a) 证明 CC 在 PP 处的法线方程为

y+px=2p3+4py + px = 2p^3 + 4p
(5)

(b) 写出 CC 在 QQ 处的法线方程。

(1)

CC 在 PP 处的法线和 CC 在 QQ 处的法线相交于点 NN

(c) 证明 NN 的坐标为

 (2(p2+pq+q2+2), −2pq(p+q))\begin{align*} &\,\big(2(p^2+pq+q^2+2),\\ &\,-2pq(p+q)\big) \end{align*}
(5)

ONON（其中 OO 为原点）垂直于 PQPQ

(d) 求 (p+q)2−3pq(p + q)^2 - 3pq 的值。

(5)

解答

(a)

解法一

思路

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对 y2=8xy^2=8x 关于 xx 求导，得到切线斜率。将点 PP 的纵坐标 y=4py=4p 代入，再取切线斜率的负倒数得到法线斜率，最后使用点斜式整理成题目要求的形式。

答题过程

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Differentiating y2=8xy^2=8x implicitly with respect to xx gives

2ydydx=8.2y\frac{\mathrm{d}y}{\mathrm{d}x}=8.

Hence

dydx=4y.\frac{\mathrm{d}y}{\mathrm{d}x}=\frac4y.

At P(2p2,4p)P(2p^2,4p), the gradient of the tangent is

44p=1p.\frac4{4p}=\frac1p.

Since p≠0p\ne0, the gradient of the normal is −p-p. Therefore, the normal at PP is

y−4p=−p(x−2p2).y-4p=-p(x-2p^2).

Expanding,

y−4p=−px+2p3.y-4p=-px+2p^3.

Adding px+4ppx+4p to both sides gives

y+px=2p3+4p,\boxed{y+px=2p^3+4p},

as required.

(b)

解法一

思路

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点 QQ 与点 PP 使用同一个参数形式，因此把 (a) 的法线方程中的 pp 换成 qq 即可。

答题过程

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Replacing pp by qq in the result from part (a), the normal at QQ is

y+qx=2q3+4q.\boxed{y+qx=2q^3+4q}.

(c)

解法一

思路

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联立两条法线方程。先相减消去 yy，利用立方差公式并使用 p≠qp\ne q 约去 p−qp-q，得到交点的横坐标；再代回其中一条法线求纵坐标。

答题过程

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At the intersection of the two normals,

y+px= 2p3+4p,y+qx= 2q3+4q.\begin{align*} y+px=&\,2p^3+4p,\\ y+qx=&\,2q^3+4q. \end{align*}

Subtracting the second equation from the first gives

 (p−q)x= 2(p3−q3)+4(p−q)= 2(p−q) ×(p2+pq+q2) +4(p−q)= 2(p−q) ×(p2+pq+q2+2).\begin{align*} &\,(p-q)x\\ =&\,2(p^3-q^3)+4(p-q)\\ =&\,2(p-q)\\ &\,\hspace{2pt}\times(p^2+pq+q^2)\\ &\,\hspace{4pt}+4(p-q)\\ =&\,2(p-q)\\ &\,\hspace{2pt}\times(p^2+pq+q^2+2). \end{align*}

Since p≠qp\ne q,

x=2(p2+pq+q2+2).x=2(p^2+pq+q^2+2).

Substituting this into the normal at PP,

 y= 2p3+4p−px= 2p3+4p −2p(p2+pq+q2+2)= 2p3+4p −2p3−2p2q −2pq2−4p= −2p2q−2pq2= −2pq(p+q).\begin{align*} &\,y\\ =&\,2p^3+4p-px\\ =&\,2p^3+4p\\ &\,\hspace{2pt}-2p(p^2+pq+q^2+2)\\ =&\,2p^3+4p\\ &\,\hspace{2pt}-2p^3-2p^2q\\ &\,\hspace{4pt}-2pq^2-4p\\ =&\,-2p^2q-2pq^2\\ =&\,-2pq(p+q). \end{align*}

Therefore,

 xN= 2(p2+pq+q2+2), yN= −2pq(p+q).\boxed{ \begin{aligned} &\,x_N\\ =&\,2(p^2+pq+q^2+2),\\ &\,y_N\\ =&\,-2pq(p+q) \end{aligned} }.

These are the required coordinates of NN.

解法二

思路

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由于题目已给出待证明的坐标，也可以直接验证该点同时满足两条法线方程。一个点同时位于两条不重合的法线上，因此就是它们的交点。

答题过程

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Let N0N_0 be the point whose coordinates satisfy

xN0= 2(p2+pq+q2+2),yN0= −2pq(p+q).\begin{align*} x_{N_0}=&\,2(p^2+pq+q^2+2),\\ y_{N_0}=&\,-2pq(p+q). \end{align*}

For the normal at PP,

 yN0+pxN0= −2pq(p+q) +2p(p2+pq+q2+2)= −2p2q−2pq2 +2p3+2p2q +2pq2+4p= 2p3+4p.\begin{align*} &\,y_{N_0}+px_{N_0}\\ =&\,-2pq(p+q)\\ &\,\hspace{2pt}+2p(p^2+pq+q^2+2)\\ =&\,-2p^2q-2pq^2\\ &\,\hspace{2pt}+2p^3+2p^2q\\ &\,\hspace{4pt}+2pq^2+4p\\ =&\,2p^3+4p. \end{align*}

Thus N0N_0 lies on the normal at PP. Similarly, for the normal at QQ,

 yN0+qxN0= −2pq(p+q) +2q(p2+pq+q2+2)= −2p2q−2pq2 +2p2q+2pq2 +2q3+4q= 2q3+4q.\begin{align*} &\,y_{N_0}+qx_{N_0}\\ =&\,-2pq(p+q)\\ &\,\hspace{2pt}+2q(p^2+pq+q^2+2)\\ =&\,-2p^2q-2pq^2\\ &\,\hspace{2pt}+2p^2q+2pq^2\\ &\,\hspace{4pt}+2q^3+4q\\ =&\,2q^3+4q. \end{align*}

Thus N0N_0 also lies on the normal at QQ. Since p≠qp\ne q, the two normals are distinct, so their intersection is

 xN= 2(p2+pq+q2+2), yN= −2pq(p+q).\boxed{ \begin{aligned} &\,x_N\\ =&\,2(p^2+pq+q^2+2),\\ &\,y_N\\ =&\,-2pq(p+q) \end{aligned} }.

解法三

思路

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把 PP 处的法线方程乘以 qq，把 QQ 处的法线方程乘以 pp，使两式的 xx 项同为 pqxpqx。相减消去 xx 后，用平方差公式和 p≠qp\ne q 求出纵坐标；再代回其中一条法线，求出横坐标。

答题过程

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The two normal equations are

y+px= 2p3+4p,y+qx= 2q3+4q.\begin{align*} y+px=&\,2p^3+4p,\\ y+qx=&\,2q^3+4q. \end{align*}

Multiplying the first by qq and the second by pp gives

 qy+pqx= 2p3q+4pq, py+pqx= 2pq3+4pq.\begin{align*} &\,qy+pqx\\ =&\,2p^3q+4pq,\\ &\,py+pqx\\ =&\,2pq^3+4pq. \end{align*}

Subtracting the first of these equations from the second,

 (p−q)y= 2pq3+4pq −2p3q−4pq= 2pq3−2p3q= −2pq(p2−q2)= −2pq(p−q) ×(p+q).\begin{align*} &\,(p-q)y\\ =&\,2pq^3+4pq\\ &\,\hspace{2pt}-2p^3q-4pq\\ =&\,2pq^3-2p^3q\\ =&\,-2pq(p^2-q^2)\\ =&\,-2pq(p-q)\\ &\,\hspace{2pt}\times(p+q). \end{align*}

Since p≠qp\ne q, dividing by p−qp-q gives

y=−2pq(p+q).y=-2pq(p+q).

Substituting into y+px=2p3+4py+px=2p^3+4p,

 px= 2p3+4p−y= 2p3+4p+2pq(p+q)= 2p3+4p +2p2q+2pq2= 2p(p2+pq+q2+2).\begin{align*} &\,px\\ =&\,2p^3+4p-y\\ =&\,2p^3+4p+2pq(p+q)\\ =&\,2p^3+4p\\ &\,\hspace{2pt}+2p^2q+2pq^2\\ =&\,2p(p^2+pq+q^2+2). \end{align*}

Since p≠0p\ne0,

x=2(p2+pq+q2+2).x=2(p^2+pq+q^2+2).

Therefore the coordinates of NN are

 xN= 2(p2+pq+q2+2), yN= −2pq(p+q).\boxed{ \begin{aligned} &\,x_N\\ =&\,2(p^2+pq+q^2+2),\\ &\,y_N\\ =&\,-2pq(p+q) \end{aligned} }.

(d)

解法一

思路

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用向量点积处理垂直条件，就不必假设直线的斜率存在。将点积因式分解后，利用 p≠qp\ne q 及一个恒正因子，可得 p+q=0p+q=0；再代入所求式，判断它是否有唯一值。

答题过程

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Using the coordinates of PP and QQ,

 PQ→= (2q2−2p2, 4q−4p)= 2(q−p)(p+q, 2).\begin{align*} &\,\overrightarrow{PQ}\\ =&\,(2q^2-2p^2,\,4q-4p)\\ =&\,2(q-p)(p+q,\,2). \end{align*}

From part (c),

 ON→= (2(p2+pq+q2+2), −2pq(p+q)).\begin{align*} &\,\overrightarrow{ON}\\ =&\,\big(2(p^2+pq+q^2+2),\\ &\,\hspace{2pt}-2pq(p+q)\big). \end{align*}

Since ON⊥PQON\perp PQ, their scalar product is zero:

 PQ→⋅ON→= 4(q−p)(p+q) ×(p2+pq+q2+2) −8pq(q−p)(p+q)= 4(q−p)(p+q) ×(p2−pq+q2+2)= 0.\begin{align*} &\,\overrightarrow{PQ}\cdot\overrightarrow{ON}\\ =&\,4(q-p)(p+q)\\ &\,\hspace{2pt}\times(p^2+pq+q^2+2)\\ &\,-8pq(q-p)(p+q)\\ =&\,4(q-p)(p+q)\\ &\,\hspace{2pt}\times(p^2-pq+q^2+2)\\ =&\,0. \end{align*}

We have q−p≠0q-p\ne0. Also,

 p2−pq+q2+2= (p−12q)2+34q2+2> 0.\begin{align*} &\,p^2-pq+q^2+2\\ =&\,\big(p-\tfrac12q\big)^2+\tfrac34q^2+2\\ >&\,0. \end{align*}

Therefore,

p+q=0,q=−p.p+q=0,\qquad q=-p.

Conversely, when q=−pq=-p, the points are

P= (2p2, 4p),Q= (2p2, −4p),N= (2(p2+2), 0).\begin{align*} P=&\,(2p^2,\,4p),\\ Q=&\,(2p^2,\,-4p),\\ N=&\,\big(2(p^2+2),\,0\big). \end{align*}

Since p≠0p\ne0, PP and QQ are distinct, and NN is not the origin. Hence PQPQ is vertical and ONON is horizontal, so they are perpendicular. Thus

 ON⊥PQ  ⟺   p+q=0.\boxed{ \begin{aligned} &\,ON\perp PQ\\ \iff&\,p+q=0 \end{aligned} }.

Substituting q=−pq=-p into the required expression,

 (p+q)2−3pq= 0−3p(−p)= 3p2.\begin{align*} &\,(p+q)^2-3pq\\ =&\,0-3p(-p)\\ =&\,3p^2. \end{align*}

The value depends on pp: for example, p=1p=1, q=−1q=-1 gives 33, whereas p=2p=2, q=−2q=-2 gives 1212. Both pairs satisfy the given conditions.

Therefore, the question as stated has no unique numerical answer.

卓老师温馨提醒

本小题的官方评分标准给出 −2-2，但其斜率法漏掉了唯一可行的情形：PQPQ 铅直、ONON 水平。

由 p≠qp\ne q，约掉 q−pq-p 本身没有问题；真正的问题是写出 mPQ=2/(p+q)m_{PQ}=2/(p+q) 时，已经排除了 p+q=0p+q=0。只有两条直线的斜率都有定义，才能用「垂直，所以斜率乘积为 −1-1」。上面的完整推导说明，题设恰好要求 p+q=0p+q=0，并非还存在另一种可用斜率法求解的情形。

也可以直接检查官方结果是否可能：

 (p+q)2−3pq= p2−pq+q2= (p−12q)2+34q2> 0,\begin{align*} &\,(p+q)^2-3pq\\ =&\,p^2-pq+q^2\\ =&\,\big(p-\tfrac12q\big)^2+\tfrac34q^2\\ >&\,0, \end{align*}

最后一步用了题设 q≠0q\ne0。因此 −2-2 不只是遗漏特殊情况后的不完整答案，而是在本题实数条件下根本不可能的值。正确结论是 q=−pq=-p，所求式为 3p23p^2，没有唯一数值答案；不要将官方的错误计算当作有效的垂直判准。