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IAL 2021 Oct FP3 Q3

A Level / Edexcel / FP3

IAL 2021 Oct Paper · Question 3

题目

Problem

The ellipse EE has equation

x264+y236=1.\frac{x^2}{64} + \frac{y^2}{36} = 1.

The line ll is the normal to EE at the point P(8cos⁡θ,6sin⁡θ)P(8\cos\theta, 6\sin\theta).

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

4xsin⁡θ−3ycos⁡θ=14sin⁡θcos⁡θ.4x\sin\theta - 3y\cos\theta = 14\sin\theta\cos\theta.

The line ll meets the xx-axis at the point AA and meets the yy-axis at the point BB. The point MM is the midpoint of ABAB.

(b) Determine a Cartesian equation for the locus of MM as θ\theta varies, giving your answer in the form ax2+by2=cax^2 + by^2 = c where aa, bb and cc are integers.

(9)
题目中文翻译

椭圆 EE 的方程为

x264+y236=1\frac{x^2}{64} + \frac{y^2}{36} = 1

直线 ll 是椭圆 EE 在点 P(8cos⁡θ,6sin⁡θ)P(8\cos\theta, 6\sin\theta) 处的法线。

(a) 利用微积分证明 ll 的方程为

4xsin⁡θ−3ycos⁡θ=14sin⁡θcos⁡θ4x\sin\theta - 3y\cos\theta = 14\sin\theta\cos\theta

直线 ll 与 xx 轴交于点 AA，与 yy 轴交于点 BB。 点 MM 是 ABAB 的中点。

(b) 求当 θ\theta 变化时，MM 的轨迹的笛卡尔方程，并写成 ax2+by2=cax^2 + by^2 = c 的形式，其中 aa、bb、cc 为整数。

解答

(a)

解法一：使用参数方程求导

思路

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由点 PP 的参数坐标求 dxdθ\frac{\mathrm{d}x}{\mathrm{d}\theta} 和 dydθ\frac{\mathrm{d}y}{\mathrm{d}\theta}，先得到切线斜率，再取负倒数得到法线斜率。把点 PP 代入点斜式并整理即可推出目标方程。

答题过程

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From

x=8cos⁡θ,y=6sin⁡θ,x=8\cos\theta, \qquad y=6\sin\theta,

we have

dxdθ=−8sin⁡θ,dydθ=6cos⁡θ.\frac{\mathrm{d}x}{\mathrm{d}\theta} =-8\sin\theta, \qquad \frac{\mathrm{d}y}{\mathrm{d}\theta} =6\cos\theta.

Thus the gradient of the tangent is

mT=dydx=−3cos⁡θ4sin⁡θ.m_T=\frac{\mathrm{d}y}{\mathrm{d}x} =-\frac{3\cos\theta}{4\sin\theta}.

Hence the gradient of the normal is

mN=−1mT=4sin⁡θ3cos⁡θ.m_N=-\frac1{m_T} =\frac{4\sin\theta}{3\cos\theta}.

The normal through PP therefore satisfies

y−6sin⁡θ= 4sin⁡θ3cos⁡θ ×(x−8cos⁡θ).\begin{align*} y-6\sin\theta =&\,\frac{4\sin\theta}{3\cos\theta} \\ &\,\hspace{4pt}\times\bigl(x-8\cos\theta\bigr). \end{align*}

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

3ycos⁡θ−18sin⁡θcos⁡θ= 4xsin⁡θ−32sin⁡θcos⁡θ,4xsin⁡θ−3ycos⁡θ= 14sin⁡θcos⁡θ.\begin{align*} 3y\cos\theta-18\sin\theta\cos\theta =&\,4x\sin\theta-32\sin\theta\cos\theta, \\ 4x\sin\theta-3y\cos\theta =&\,14\sin\theta\cos\theta. \end{align*}

Therefore,

4xsin⁡θ−3ycos⁡θ=14sin⁡θcos⁡θ.\boxed{ 4x\sin\theta-3y\cos\theta =14\sin\theta\cos\theta }.

The limiting horizontal and vertical normals are also represented by this equation.

解法二：隐式求导

思路

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直接对椭圆方程隐式求导，得到切线斜率的 x,yx,y 表达式。代入 PP 的参数坐标后求法线斜率，再按点斜式整理成指定方程。

答题过程

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Differentiating

x264+y236=1\frac{x^2}{64}+\frac{y^2}{36}=1

implicitly gives

x32+y18dydx=0.\frac{x}{32}+\frac{y}{18} \frac{\mathrm{d}y}{\mathrm{d}x}=0.

Therefore,

dydx=−9x16y.\frac{\mathrm{d}y}{\mathrm{d}x} =-\frac{9x}{16y}.

At P=(8cos⁡θ,6sin⁡θ)P=(8\cos\theta,6\sin\theta),

mT=−3cos⁡θ4sin⁡θ,mN=4sin⁡θ3cos⁡θ.m_T=-\frac{3\cos\theta}{4\sin\theta}, \qquad m_N=\frac{4\sin\theta}{3\cos\theta}.

Hence

y−6sin⁡θ= 4sin⁡θ3cos⁡θ ×(x−8cos⁡θ),4xsin⁡θ−3ycos⁡θ= 14sin⁡θcos⁡θ.\begin{align*} y-6\sin\theta =&\,\frac{4\sin\theta}{3\cos\theta} \\ &\,\hspace{4pt}\times\bigl(x-8\cos\theta\bigr), \\ 4x\sin\theta-3y\cos\theta =&\,14\sin\theta\cos\theta. \end{align*}

Thus

4xsin⁡θ−3ycos⁡θ=14sin⁡θcos⁡θ.\boxed{ 4x\sin\theta-3y\cos\theta =14\sin\theta\cos\theta }.

(b)

解法一

思路

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分别令法线方程中的 y=0y=0、x=0x=0，求出两个截距点 A,BA,B。由中点公式得到 MM 的参数坐标，再以 sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1 消去参数。

答题过程

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Setting y=0y=0 in the equation of ll gives

A=(72cos⁡θ,0).A=\bigg(\frac72\cos\theta,0\bigg).

Setting x=0x=0 gives

B=(0,−143sin⁡θ).B=\bigg(0,-\frac{14}{3}\sin\theta\bigg).

Therefore, the midpoint of ABAB is

M=(74cos⁡θ,−73sin⁡θ).M=\bigg(\frac74\cos\theta,-\frac73\sin\theta\bigg).

Let the coordinates of MM be (x,y)(x,y). Then

cos⁡θ=4x7,sin⁡θ=−3y7.\cos\theta=\frac{4x}{7}, \qquad \sin\theta=-\frac{3y}{7}.

Using sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1,

(4x7)2+(3y7)2= 1,16x2+9y2= 49.\begin{align*} \bigg(\frac{4x}{7}\bigg)^2 +\bigg(\frac{3y}{7}\bigg)^2 =&\,1, \\ 16x^2+9y^2 =&\,49. \end{align*}

Hence the Cartesian equation of the locus is

16x2+9y2=49.\boxed{16x^2+9y^2=49}.