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FP3 2024 June Q6

A Level / Edexcel / FP3

IAL 2024 June Paper · Question 6

题目

Problem

6.

The ellipse EE has equation

x225+y29=1\frac{x^2}{25}+\frac{y^2}{9}=1

The line ll is the normal to EE at the point P(5cos⁡θ,3sin⁡θ)P(5\cos\theta,3\sin\theta) where 0<θ<π20<\theta<\frac\pi2

(a) Using calculus, show that an equation for ll is

5xsin⁡θ−3ycos⁡θ=16sin⁡θcos⁡θ5x\sin\theta-3y\cos\theta=16\sin\theta\cos\theta
(4)

Given that

• ll intersects the yy-axis at the point QQ

• the midpoint of the line segment PQPQ is MM

(b) determine the exact maximum area of triangle OMPOMP as θ\theta varies, where OO is the origin.

You must justify your answer.

(5)
题目中文翻译

6.

椭圆 EE 的方程为

x225+y29=1\frac{x^2}{25}+\frac{y^2}{9}=1

直线 ll 是 EE 在点 P(5cos⁡θ,3sin⁡θ)P(5\cos\theta,3\sin\theta) 处的法线，其中 0<θ<π20<\theta<\frac\pi2

(a) 用微积分证明，ll 的方程为

5xsin⁡θ−3ycos⁡θ=16sin⁡θcos⁡θ5x\sin\theta-3y\cos\theta=16\sin\theta\cos\theta
(4)

已知

• ll 与 yy 轴交于点 QQ

• 线段 PQPQ 的中点为 MM

(b) 当 θ\theta 变化时，求三角形 OMPOMP 面积的精确最大值，其中 OO 为原点。

你必须给出证明。

(5)

解答

(a)

解法一

思路

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由点 PP 的参数坐标分别求 dx/dθ\mathrm{d}x/\mathrm{d}\theta 与 dy/dθ\mathrm{d}y/\mathrm{d}\theta，从而得到切线斜率。法线斜率是切线斜率的负倒数；把 PP 代入点斜式后整理，即可自然推出题设方程。

答题过程

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At the point PP,

x=5cos⁡θ,y=3sin⁡θ.x=5\cos\theta, \qquad y=3\sin\theta.

Therefore,

dxdθ=−5sin⁡θ,dydθ=3cos⁡θ.\frac{\mathrm{d}x}{\mathrm{d}\theta} =-5\sin\theta, \qquad \frac{\mathrm{d}y}{\mathrm{d}\theta} =3\cos\theta.

The gradient of the tangent is

dydx=dy/dθdx/dθ=−3cos⁡θ5sin⁡θ.\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{\mathrm{d}y/\mathrm{d}\theta} {\mathrm{d}x/\mathrm{d}\theta} =-\frac{3\cos\theta}{5\sin\theta}.

Hence the gradient of the normal is

ml=5sin⁡θ3cos⁡θ.m_l=\frac{5\sin\theta}{3\cos\theta}.

Using the point P(5cos⁡θ,3sin⁡θ)P(5\cos\theta,3\sin\theta), an equation of the normal is

y−3sin⁡θ=5sin⁡θ3cos⁡θ(x−5cos⁡θ).y-3\sin\theta =\frac{5\sin\theta}{3\cos\theta} \bigl(x-5\cos\theta\bigr).

Multiplying by 3cos⁡θ3\cos\theta and rearranging,

3ycos⁡θ−9sin⁡θcos⁡θ= 5xsin⁡θ−25sin⁡θcos⁡θ,5xsin⁡θ−3ycos⁡θ= 16sin⁡θcos⁡θ.\begin{align*} 3y\cos\theta-9\sin\theta\cos\theta =&\,5x\sin\theta-25\sin\theta\cos\theta, \\ 5x\sin\theta-3y\cos\theta =&\,16\sin\theta\cos\theta. \end{align*}

Thus an equation of ll is

5xsin⁡θ−3ycos⁡θ=16sin⁡θcos⁡θ.\boxed{ 5x\sin\theta-3y\cos\theta =16\sin\theta\cos\theta }.

(b)

解法一：求中点坐标并使用行列式面积

思路

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先由 (a) 的法线方程求 QQ 的坐标，再求中点 MM。以 OO 为原点时，三角形面积可以直接由向量 OM→\overrightarrow{OM}、OP→\overrightarrow{OP} 的二维行列式求得。最后把面积化为 103sin⁡2θ\frac{10}{3}\sin2\theta，即可严格判断最大值。

答题过程

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At QQ, x=0x=0. Using the equation from part (a),

−3yQcos⁡θ=16sin⁡θcos⁡θ.-3y_Q\cos\theta =16\sin\theta\cos\theta.

Since 0<θ<π/20<\theta<\pi/2, cos⁡θ≠0\cos\theta\ne0, so

Q=(0,−163sin⁡θ).Q=\biggl(0,-\frac{16}{3}\sin\theta\biggr).

As MM is the midpoint of PQPQ,

M= (5cos⁡θ+02,3sin⁡θ−163sin⁡θ2)= (52cos⁡θ,−76sin⁡θ).\begin{align*} M =&\,\biggl( \frac{5\cos\theta+0}{2}, \frac{3\sin\theta-\frac{16}{3}\sin\theta}{2} \biggr) \\ =&\,\biggl( \frac52\cos\theta, -\frac76\sin\theta \biggr). \end{align*}

Hence

Area⁡(△OMP)= 12∣(52cos⁡θ)(3sin⁡θ) −(−76sin⁡θ)(5cos⁡θ)∣= 12(152+356)sin⁡θcos⁡θ= 203sin⁡θcos⁡θ= 103sin⁡2θ.\begin{align*} \operatorname{Area}(\triangle OMP) =&\,\frac12\Bigg| \biggl(\frac52\cos\theta\biggr) \bigl(3\sin\theta\bigr) \\ &\,\hspace{4pt} -\biggl(-\frac76\sin\theta\biggr) \bigl(5\cos\theta\bigr) \Bigg| \\ =&\,\frac12\biggl( \frac{15}{2}+\frac{35}{6} \biggr)\sin\theta\cos\theta \\ =&\,\frac{20}{3}\sin\theta\cos\theta \\ =&\,\frac{10}{3}\sin2\theta. \end{align*}

For 0<θ<π/20<\theta<\pi/2, we have 0<2θ<π0<2\theta<\pi and sin⁡2θ⩽1\sin2\theta\leqslant1, with equality when θ=π/4\theta=\pi/4. Therefore,

max⁡Area⁡(△OMP)=103.\boxed{ \max\operatorname{Area}(\triangle OMP)=\frac{10}{3} }.

解法二：利用中点产生的面积比例

思路

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因为 MM 是 PQPQ 的中点，△OPM\triangle OPM 与 △OPQ\triangle OPQ 对应于同一直线 PQPQ 上的底边分别为 PMPM、PQPQ，且从 OO 到该直线的高相同，所以前者面积恰为后者的一半。这样无需显式求出 MM 的坐标。

答题过程

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From part (a),

Q=(0,−163sin⁡θ).Q=\biggl(0,-\frac{16}{3}\sin\theta\biggr).

Since MM is the midpoint of PQPQ, PM=12PQPM=\frac12PQ. The triangles OPMOPM and OPQOPQ have the same perpendicular height from OO to the line PQPQ. Therefore,

Area⁡(△OPM)=12Area⁡(△OPQ).\operatorname{Area}(\triangle OPM) =\frac12\operatorname{Area}(\triangle OPQ).

Using the coordinates of PP and QQ,

Area⁡(△OPM)= 12⋅12∣5cos⁡θ(−163sin⁡θ)∣= 203sin⁡θcos⁡θ= 103sin⁡2θ.\begin{align*} \operatorname{Area}(\triangle OPM) =&\,\frac12\cdot\frac12 \Bigg| 5\cos\theta \biggl(-\frac{16}{3}\sin\theta\biggr) \Bigg| \\ =&\,\frac{20}{3}\sin\theta\cos\theta \\ =&\,\frac{10}{3}\sin2\theta. \end{align*}

Because sin⁡2θ⩽1\sin2\theta\leqslant1, with equality at θ=π/4\theta=\pi/4 in the given interval,

max⁡Area⁡(△OMP)=103.\boxed{ \max\operatorname{Area}(\triangle OMP)=\frac{10}{3} }.