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IAL 2021 June FP1 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))(2(p^2 + pq + q^2 + 2), \ -2pq(p + q))

(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

(b) 写出 CC 在 QQ 处的法线方程。

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))(2(p^2 + pq + q^2 + 2), \ -2pq(p + q))

ONON（其中 OO 为原点）垂直于 PQPQ

(d) 求 (p+q)2−3pq(p + q)^2 - 3pq 的值。

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 and rearranging 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)(p^2+pq+q^2)\\ &\,\hspace{2pt}+4(p-q)\\ =&\,2(p-q)(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)= −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^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)= 2p3+4p.\begin{align*} y_{N_0}+px_{N_0} =&\,-2pq(p+q)\\ &\,\hspace{2pt}+2p(p^2+pq+q^2+2)\\ =&\,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)= 2q3+4q.\begin{align*} y_{N_0}+qx_{N_0} =&\,-2pq(p+q)\\ &\,\hspace{2pt}+2q(p^2+pq+q^2+2)\\ =&\,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} }.

(d)

解法一

思路

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按官方评分资料的预期路线，分别用 P,QP,Q 与 (c) 的点 NN 求 PQPQ、ONON 的斜率，再利用垂直直线斜率乘积为 −1-1。这一做法隐含 p+q≠0p+q\ne0。原题没有排除 q=−pq=-p；在该特例中 PQPQ 竖直而 ONON 水平，垂直条件仍成立，但所求式不唯一。因此下文给出的是官方预期的非竖直情形。

答题过程

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For the intended non-vertical case,

mPQ= 4q−4p2q2−2p2= 4(q−p)2(q−p)(p+q)= 2p+q.\begin{align*} m_{PQ} =&\,\frac{4q-4p}{2q^2-2p^2}\\ =&\,\frac{4(q-p)}{2(q-p)(p+q)}\\ =&\,\frac2{p+q}. \end{align*}

Using the coordinates of NN from part (c),

mON= −2pq(p+q)2(p2+pq+q2+2)= −pq(p+q)p2+pq+q2+2.\begin{align*} m_{ON} =&\,\frac{-2pq(p+q)} {2(p^2+pq+q^2+2)}\\ =&\,-\frac{pq(p+q)} {p^2+pq+q^2+2}. \end{align*}

Since ONON is perpendicular to PQPQ,

mPQmON= −1,= 2p+q ×(−pq(p+q)p2+pq+q2+2)= −1,−2pqp2+pq+q2+2= −1.\begin{align*} m_{PQ}m_{ON}=&\,-1,\\ =&\,\frac2{p+q}\\ &\,\hspace{2pt}\times \bigg(-\frac{pq(p+q)}{p^2+pq+q^2+2}\bigg)\\ =&\,-1,\\ \frac{-2pq}{p^2+pq+q^2+2} =&\,-1. \end{align*}

Therefore,

p2+pq+q2+2= 2pq,p2−pq+q2= −2.\begin{align*} p^2+pq+q^2+2=&\,2pq,\\ p^2-pq+q^2=&\,-2. \end{align*}

Finally,

(p+q)2−3pq= p2−pq+q2= −2.\begin{align*} (p+q)^2-3pq =&\,p^2-pq+q^2\\ =&\,\boxed{-2}. \end{align*}