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IAL 2021 Jan FP3 Q9

A Level / Edexcel / FP3

IAL 2021 Jan Paper · Question 9

题目

Problem

The ellipse EE has equation

x225+y216=1.\frac{x^2}{25} + \frac{y^2}{16} = 1.

The point PP lies on the ellipse and has coordinates (5cos⁡θ,4sin⁡θ)(5\cos\theta, 4\sin\theta) where 0<θ<π20 < \theta < \frac{\pi}{2}.

The line ll is the normal to the ellipse at the point PP.

(a) Show that an equation for ll is

5xsin⁡θ−4ycos⁡θ=9sin⁡θcos⁡θ.5x\sin\theta - 4y\cos\theta = 9\sin\theta\cos\theta.

The point FF is the focus of EE that lies on the positive xx-axis.

(b) Determine the coordinates of FF.

The line ll crosses the xx-axis at the point QQ.

(c) Show that

QFPF=e\frac{QF}{PF} = e

where ee is the eccentricity of EE.

(12)
题目中文翻译

椭圆 EE 的方程为

x225+y216=1\frac{x^2}{25} + \frac{y^2}{16} = 1

点 PP 在椭圆上，坐标为 (5cos⁡θ,4sin⁡θ)(5\cos\theta, 4\sin\theta)，其中 0<θ<π20 < \theta < \frac{\pi}{2}。

直线 ll 是点 PP 处的法线。

(a) 证明 ll 的方程为

5xsin⁡θ−4ycos⁡θ=9sin⁡θcos⁡θ5x\sin\theta - 4y\cos\theta = 9\sin\theta\cos\theta

点 FF 是椭圆 EE 在正 xx 轴上的焦点。

(b) 求 FF 的坐标。

直线 ll 与 xx 轴交于点 QQ。

(c) 证明

QFPF=e\frac{QF}{PF} = e

其中 ee 为 EE 的离心率。

解答

(a)

解法一

思路

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用参数方程分别求 dx/dθ\mathrm{d}x/\mathrm{d}\theta 与 dy/dθ\mathrm{d}y/\mathrm{d}\theta，得到切线斜率，再取负倒数求法线斜率。随后使用点斜式并整理到题目给定形式。

答题过程

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For P=(5cos⁡θ,4sin⁡θ)P=(5\cos\theta,4\sin\theta),

dxdθ=−5sin⁡θ,dydθ=4cos⁡θ.\frac{\mathrm{d}x}{\mathrm{d}\theta}=-5\sin\theta, \qquad \frac{\mathrm{d}y}{\mathrm{d}\theta}=4\cos\theta.

Thus the gradient of the tangent is

dydx=4cos⁡θ−5sin⁡θ.\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{4\cos\theta}{-5\sin\theta}.

Since 0<θ<π20<\theta<\frac\pi2, both sin⁡θ\sin\theta and cos⁡θ\cos\theta are non-zero. Therefore the gradient of the normal is

mN=5sin⁡θ4cos⁡θ.m_N=\frac{5\sin\theta}{4\cos\theta}.

The normal through PP has equation

y−4sin⁡θ=5sin⁡θ4cos⁡θ(x−5cos⁡θ).y-4\sin\theta =\frac{5\sin\theta}{4\cos\theta} (x-5\cos\theta).

Multiplying by 4cos⁡θ4\cos\theta and rearranging,

4ycos⁡θ−16sin⁡θcos⁡θ= 5xsin⁡θ−25sin⁡θcos⁡θ,5xsin⁡θ−4ycos⁡θ= 9sin⁡θcos⁡θ.\begin{align*} 4y\cos\theta-16\sin\theta\cos\theta =&\,5x\sin\theta-25\sin\theta\cos\theta,\\[4mm] 5x\sin\theta-4y\cos\theta =&\,9\sin\theta\cos\theta. \end{align*}

Hence

5xsin⁡θ−4ycos⁡θ=9sin⁡θcos⁡θ.\boxed{5x\sin\theta-4y\cos\theta =9\sin\theta\cos\theta}.

(b)

解法一

思路

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椭圆的半长轴为 a=5a=5、半短轴为 b=4b=4。焦距满足 c2=a2−b2c^2=a^2-b^2，正 xx 轴上的焦点即为 (c,0)(c,0)。

答题过程

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For the ellipse,

a=5,b=4.a=5, \qquad b=4.

Hence the focal distance is

c=a2−b2=25−16=3.c=\sqrt{a^2-b^2} =\sqrt{25-16}=3.

Therefore the focus on the positive xx-axis is

F=(3,0).\boxed{F=(3,0)}.

Also, the eccentricity is

e=ca=35.e=\frac ca=\frac35.

(c)

解法一

思路

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先在 (a) 的法线方程中令 y=0y=0 求出 QQ。然后分别求 QFQF 和 PFPF；计算 PFPF 时利用 sin⁡2θ=1−cos⁡2θ\sin^2\theta=1-\cos^2\theta，所得二次式恰为完全平方。给定角度范围可确定长度表达式的正号。

答题过程

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

5xsin⁡θ=9sin⁡θcos⁡θ.5x\sin\theta=9\sin\theta\cos\theta.

Since sin⁡θ>0\sin\theta>0,

x=95cos⁡θ,x=\frac95\cos\theta,

so

Q=(95cos⁡θ,0).Q=\left(\frac95\cos\theta,0\right).

As 0<cos⁡θ<10<\cos\theta<1,

QF=3−95cos⁡θ=35(5−3cos⁡θ).QF=3-\frac95\cos\theta =\frac35(5-3\cos\theta).

Also,

PF2= (5cos⁡θ−3)2+(4sin⁡θ)2= 25cos⁡2θ−30cos⁡θ+9 +16(1−cos⁡2θ)= 9cos⁡2θ−30cos⁡θ+25= (5−3cos⁡θ)2.\begin{align*} PF^2 =&\,(5\cos\theta-3)^2+(4\sin\theta)^2\\[4mm] =&\,25\cos^2\theta-30\cos\theta+9\\[4mm] &\,\hspace{2pt}+16(1-\cos^2\theta)\\[4mm] =&\,9\cos^2\theta-30\cos\theta+25\\[4mm] =&\,(5-3\cos\theta)^2. \end{align*}

Since 5−3cos⁡θ>05-3\cos\theta>0,

PF=5−3cos⁡θ.PF=5-3\cos\theta.

Therefore

QFPF=35(5−3cos⁡θ)5−3cos⁡θ=35=e.\frac{QF}{PF} =\frac{\frac35(5-3\cos\theta)} {5-3\cos\theta} =\frac35=e.

Hence

QFPF=e.\boxed{\frac{QF}{PF}=e}.