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IAL 2025 Jan FP3 Q7

A Level / Edexcel / FP3

IAL 2025 Jan Paper · Question 7

题目

Problem

The ellipse EE has equation

x2a2+y2b2=1a>b>0\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \qquad a>b>0

The point P(acos⁡θ,bsin⁡θ)P(a\cos\theta, b\sin\theta) lies on E where 0<θ<π20<\theta<\frac{\pi}{2}

(a) Use calculus to show that an equation of the normal to E at P is

by=axtan⁡θ+(b2−a2)sin⁡θby = ax\tan\theta + (b^2-a^2)\sin\theta

The normal to E at P meets E again on the y-axis at the point B.

Given that O is the origin and that the area of triangle OBP is 3b24\dfrac{3b^2}{4}

(b) show that sin⁡θ=12\sin\theta = \dfrac{1}{2}

(c) determine, in terms of aa only, the exact coordinates of the point P.

(4)
(5)
(3)
题目中文翻译

椭圆 EE 的方程为

x2a2+y2b2=1a>b>0\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \qquad a>b>0

点 P(acos⁡θ,bsin⁡θ)P(a\cos\theta, b\sin\theta) 在 E 上，其中 0<θ<π20<\theta<\frac{\pi}{2}。

(a) 用微积分证明，P 点处的法线方程为

by=axtan⁡θ+(b2−a2)sin⁡θby = ax\tan\theta + (b^2-a^2)\sin\theta

P 点处的法线与 E 再次相交于 y 轴上的点 B。

已知 O 为原点，三角形 OBP 的面积为 3b24\dfrac{3b^2}{4}。

(b) 证明 sin⁡θ=12\sin\theta = \dfrac{1}{2}。

(c) 用仅含 aa 的形式求点 P 的准确坐标。

解答

(a)

解法一

思路

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把椭圆上的点视为参数曲线，分别对 x=acos⁡θx=a\cos\theta、y=bsin⁡θy=b\sin\theta 关于 θ\theta 求导，从而求出切线斜率。法线斜率是切线斜率的负倒数，再用点斜式并整理成题目指定的形式。

答题过程

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Using the parametric coordinates of PP,

dxdθ= −asin⁡θ,dydθ= bcos⁡θ.\begin{align*} \frac{\mathrm dx}{\mathrm d\theta} =&\,-a\sin\theta,\\ \frac{\mathrm dy}{\mathrm d\theta} =&\,b\cos\theta. \end{align*}

Therefore, the gradient of the tangent is

dydx=−bcos⁡θasin⁡θ,\frac{\mathrm dy}{\mathrm dx} =-\frac{b\cos\theta}{a\sin\theta},

so the gradient of the normal is

mN=asin⁡θbcos⁡θ.m_N=\frac{a\sin\theta}{b\cos\theta}.

The normal through P=(acos⁡θ,bsin⁡θ)P=(a\cos\theta,b\sin\theta) is therefore

y−bsin⁡θ=asin⁡θbcos⁡θ(x−acos⁡θ).y-b\sin\theta =\frac{a\sin\theta}{b\cos\theta} (x-a\cos\theta).

Multiplying by bcos⁡θb\cos\theta gives

bycos⁡θ−b2sin⁡θcos⁡θ= axsin⁡θ −a2sin⁡θcos⁡θ.\begin{align*} by\cos\theta-b^2\sin\theta\cos\theta =&\,ax\sin\theta\\ &\,\hspace{2pt}-a^2\sin\theta\cos\theta. \end{align*}

Since 0<θ<π20<\theta<\frac\pi2, cos⁡θ>0\cos\theta>0. Dividing by cos⁡θ\cos\theta and rearranging,

by=axtan⁡θ+(b2−a2)sin⁡θ.\boxed{ by=ax\tan\theta+(b^2-a^2)\sin\theta}.

解法二

思路

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官方评分资料也允许对椭圆方程隐式求导。先得到含 x,yx,y 的切线斜率，再代入点 PP，便会得到与参数求导相同的法线斜率。

答题过程

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Differentiating the equation of the ellipse implicitly,

2xa2+2yb2dydx= 0,dydx= −b2xa2y.\begin{align*} \frac{2x}{a^2} +\frac{2y}{b^2}\frac{\mathrm dy}{\mathrm dx} =&\,0,\\ \frac{\mathrm dy}{\mathrm dx} =&\,-\frac{b^2x}{a^2y}. \end{align*}

At P=(acos⁡θ,bsin⁡θ)P=(a\cos\theta,b\sin\theta),

dydx=−bcos⁡θasin⁡θ.\frac{\mathrm dy}{\mathrm dx} =-\frac{b\cos\theta}{a\sin\theta}.

Hence the gradient of the normal is

mN=asin⁡θbcos⁡θ.m_N=\frac{a\sin\theta}{b\cos\theta}.

Using the point-slope form,

y−bsin⁡θ=asin⁡θbcos⁡θ(x−acos⁡θ).y-b\sin\theta =\frac{a\sin\theta}{b\cos\theta} (x-a\cos\theta).

Thus

by−b2sin⁡θ= axtan⁡θ−a2sin⁡θ,by= axtan⁡θ+(b2−a2)sin⁡θ.\begin{align*} by-b^2\sin\theta =&\,ax\tan\theta-a^2\sin\theta,\\ by=&\,ax\tan\theta +(b^2-a^2)\sin\theta. \end{align*}

Therefore,

by=axtan⁡θ+(b2−a2)sin⁡θ.\boxed{ by=ax\tan\theta+(b^2-a^2)\sin\theta}.

(b)

解法一

思路

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先利用 BB 同时在椭圆和 yy 轴上，判断其坐标是 (0,−b)(0,-b)，再把它代入 (a) 的法线方程。另一方面，以 OBOB 为底、点 PP 到 yy 轴的距离为高，写出三角形面积。联立两式并令 s=sin⁡θs=\sin\theta，即可得到只含 ss 的三次方程。

答题过程

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Since BB lies on the yy-axis and on the ellipse, B=(0,±b)B=(0,\pm b). In the normal equation, setting x=0x=0 gives

by=(b2−a2)sin⁡θ<0,by=(b^2-a^2)\sin\theta<0,

because a>b>0a>b>0 and 0<θ<π20<\theta<\frac\pi2. Hence

B=(0,−b).B=(0,-b).

Substituting BB into the normal equation gives

−b2=(b2−a2)sin⁡θ.-b^2=(b^2-a^2)\sin\theta.

Also, the base OBOB has length bb, and the perpendicular height of PP from the yy-axis is acos⁡θa\cos\theta. Therefore,

12abcos⁡θ= 3b24,2acos⁡θ= 3b,b2= 4a29cos⁡2θ.\begin{align*} \frac12ab\cos\theta=&\,\frac{3b^2}{4},\\ 2a\cos\theta=&\,3b,\\ b^2=&\,\frac{4a^2}{9}\cos^2\theta. \end{align*}

Let s=sin⁡θs=\sin\theta. Then

b2=4a29(1−s2).b^2=\frac{4a^2}{9}(1-s^2).

Substituting this into the equation from the normal,

4(1−s2)=[9−4(1−s2)]s.4(1-s^2) =\big[9-4(1-s^2)\big]s.

Therefore,

4−4s2= 5s+4s3,4s3+4s2+5s−4= 0,(2s−1)(2s2+3s+4)= 0.\begin{align*} 4-4s^2=&\,5s+4s^3,\\ 4s^3+4s^2+5s-4=&\,0,\\ (2s-1)(2s^2+3s+4)=&\,0. \end{align*}

The quadratic factor has discriminant

32−4(2)(4)=−23<0,3^2-4(2)(4)=-23<0,

so it has no real roots. Hence 2s−1=02s-1=0, and therefore

sin⁡θ=12.\boxed{\sin\theta=\frac12}.

解法二

思路

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官方替代路线不立即消去 a,ba,b。先分别由法线条件和面积条件写出 sin⁡θ\sin\theta、cos⁡θ\cos\theta，再代入 sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1，得到关于 a2,b2a^2,b^2 的多项式并因式分解。

答题过程

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From the normal equation at B=(0,−b)B=(0,-b),

sin⁡θ=b2a2−b2.\sin\theta=\frac{b^2}{a^2-b^2}.

From the area condition,

cos⁡θ=3b2a.\cos\theta=\frac{3b}{2a}.

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

b4(a2−b2)2+9b24a2=1.\frac{b^4}{(a^2-b^2)^2} +\frac{9b^2}{4a^2}=1.

Multiplying by 4a2(a2−b2)24a^2(a^2-b^2)^2 and expanding gives

 9b6−18a2b4+17a4b2−4a6=0, (3b2−a2)(3b4−5a2b2+4a4)=0.\begin{align*} &\,9b^6-18a^2b^4+17a^4b^2-4a^6=0,\\ &\,(3b^2-a^2) (3b^4-5a^2b^2+4a^4)=0. \end{align*}

Treating the second factor as a quadratic in b2b^2, its discriminant is

(−5a2)2−4(3)(4a4)=−23a4<0.(-5a^2)^2-4(3)(4a^4)=-23a^4<0.

Therefore, the second factor cannot be zero for real a,ba,b, so

3b2−a2=0.3b^2-a^2=0.

Hence

sin⁡θ= b2a2−b2= b23b2−b2= 12.\begin{align*} \sin\theta =&\,\frac{b^2}{a^2-b^2}\\ =&\,\frac{b^2}{3b^2-b^2}\\ =&\,\boxed{\frac12}. \end{align*}

(c)

解法一

思路

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承接 (b)，由角度范围确定 cos⁡θ\cos\theta 取正值。再利用面积条件 2acos⁡θ=3b2a\cos\theta=3b 把 bb 写成 aa，最后代入 P=(acos⁡θ,bsin⁡θ)P=(a\cos\theta,b\sin\theta)。

答题过程

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Since 0<θ<π20<\theta<\frac\pi2 and sin⁡θ=12\sin\theta=\frac12,

cos⁡θ=32.\cos\theta=\frac{\sqrt3}{2}.

From the area condition in part (b),

b= 2a3cos⁡θ= 33a.\begin{align*} b =&\,\frac{2a}{3}\cos\theta\\ =&\,\frac{\sqrt3}{3}a. \end{align*}

Therefore,

P= (acos⁡θ,bsin⁡θ)= (32a,36a).\begin{align*} P =&\,(a\cos\theta,b\sin\theta)\\ =&\,\left( \frac{\sqrt3}{2}a, \frac{\sqrt3}{6}a \right). \end{align*}

Hence the exact coordinates are

P(3a2,3a6).\boxed{ P\left(\frac{\sqrt3a}{2},\frac{\sqrt3a}{6}\right)}.