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IAL 2026 Jan FP3 Q8

A Level / Edexcel / FP3

IAL 2026 Jan Paper · Question 8

题目

Problem

The ellipse EE has equation

x216+y29=1\frac{x^2}{16} + \frac{y^2}{9} = 1

The point P(4cos⁡θ,3sin⁡θ)P(4\cos \theta, 3\sin \theta) lies on EE where 0<θ<π20 < \theta < \frac{\pi}{2}

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

4xsin⁡θ−3ycos⁡θ=7sin⁡θcos⁡θ4x\sin\theta - 3y\cos\theta = 7\sin\theta\cos\theta

The normal to E at P meets the x-axis at the point A.

(b) Show that the area of triangle OAP, where O is the origin, is ksin⁡2θk\sin 2\theta, where kk is a rational number to be determined.

(c) Hence state the value of the maximum area for triangle OAP.

(4)
(4)
(1)
题目中文翻译

椭圆 EE 的方程为

x216+y29=1\frac{x^2}{16} + \frac{y^2}{9} = 1

点 P(4cos⁡θ,3sin⁡θ)P(4\cos\theta, 3\sin\theta) 在 EE 上，其中 0<θ<π20 < \theta < \frac{\pi}{2}。

(a) 用微积分证明，PP 点处的法线方程为

4xsin⁡θ−3ycos⁡θ=7sin⁡θcos⁡θ4x\sin\theta - 3y\cos\theta = 7\sin\theta\cos\theta

椭圆 EE 在 PP 点处的法线与 x 轴交于点 A。

(b) 证明三角形 OAP 的面积为 ksin⁡2θk\sin 2\theta，其中 kk 为待确定的有理数。

(c) 由此给出三角形 OAP 的最大面积。

解答

(a)

解法一

思路

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先对椭圆方程隐式求导，代入 PP 的坐标得到切线斜率，再取负倒数求法线斜率。将点斜式整理后应自然得到题目指定的方程。

答题过程

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Differentiating the equation of the ellipse implicitly,

x8+2y9dydx=0.\frac{x}{8} +\frac{2y}{9}\frac{\mathrm{d}y}{\mathrm{d}x}=0.

Therefore,

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

At P(4cos⁡θ,3sin⁡θ)P(4\cos\theta,3\sin\theta), the gradient of the tangent is

mT= −9(4cos⁡θ)16(3sin⁡θ)= −3cos⁡θ4sin⁡θ.\begin{align*} m_T =&\,-\frac{9(4\cos\theta)} {16(3\sin\theta)}\\ =&\,-\frac{3\cos\theta}{4\sin\theta}. \end{align*}

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 equation of the normal at PP is therefore

y−3sin⁡θ=4sin⁡θ3cos⁡θ(x−4cos⁡θ).y-3\sin\theta =\frac{4\sin\theta}{3\cos\theta} \big(x-4\cos\theta\big).

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

3ycos⁡θ−9sin⁡θcos⁡θ= 4xsin⁡θ−16sin⁡θcos⁡θ,4xsin⁡θ−3ycos⁡θ= 7sin⁡θcos⁡θ,\begin{align*} 3y\cos\theta-9\sin\theta\cos\theta =&\,4x\sin\theta-16\sin\theta\cos\theta,\\ 4x\sin\theta-3y\cos\theta =&\,7\sin\theta\cos\theta, \end{align*}

as required.

解法二

思路

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官方评分资料也允许直接利用点 PP 的参数表示。分别求 x,yx,y 关于 θ\theta 的导数，再用 dy/dx=(dy/dθ)/(dx/dθ)\mathrm{d}y/\mathrm{d}x=(\mathrm{d}y/\mathrm{d}\theta)/(\mathrm{d}x/\mathrm{d}\theta)，能更快得到同一个切线斜率；后续法线方程与解法一相同。

答题过程

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From

x=4cos⁡θx=4\cos\theta

and

y=3sin⁡θ,y=3\sin\theta,

we obtain

dxdθ=−4sin⁡θ\frac{\mathrm{d}x}{\mathrm{d}\theta} =-4\sin\theta

and

dydθ=3cos⁡θ.\frac{\mathrm{d}y}{\mathrm{d}\theta} =3\cos\theta.

Thus

dydx= dy/dθdx/dθ= −3cos⁡θ4sin⁡θ.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\frac{\mathrm{d}y/\mathrm{d}\theta} {\mathrm{d}x/\mathrm{d}\theta}\\ =&\,-\frac{3\cos\theta}{4\sin\theta}. \end{align*}

Hence the normal has gradient

4sin⁡θ3cos⁡θ.\frac{4\sin\theta}{3\cos\theta}.

Using the point PP,

y−3sin⁡θ=4sin⁡θ3cos⁡θ(x−4cos⁡θ).y-3\sin\theta =\frac{4\sin\theta}{3\cos\theta} \big(x-4\cos\theta\big).

Therefore,

4xsin⁡θ−3ycos⁡θ=7sin⁡θcos⁡θ.\boxed{ 4x\sin\theta-3y\cos\theta =7\sin\theta\cos\theta}.

(b)

解法一

思路

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点 AA 在 xx 轴上，所以在 (a) 的法线方程中令 y=0y=0，求出其横坐标。再以 OAOA 为底、PP 的纵坐标为高求三角形面积，并用 sin⁡2θ=2sin⁡θcos⁡θ\sin2\theta=2\sin\theta\cos\theta 化为指定形式。

答题过程

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At AA, y=0y=0. Therefore, using the normal equation from part (a),

4xAsin⁡θ=7sin⁡θcos⁡θ.4x_A\sin\theta =7\sin\theta\cos\theta.

Since 0<θ<π20<\theta<\frac{\pi}{2}, sin⁡θ>0\sin\theta>0, so

xA=74cos⁡θ.x_A=\frac74\cos\theta.

Also, the perpendicular height of PP above the xx-axis is 3sin⁡θ3\sin\theta. Hence

Area⁡(OAP)= 12(74cos⁡θ)(3sin⁡θ)= 218sin⁡θcos⁡θ= 2116sin⁡2θ.\begin{align*} \operatorname{Area}(OAP) =&\,\frac12\left(\frac74\cos\theta\right) (3\sin\theta)\\ =&\,\frac{21}{8}\sin\theta\cos\theta\\ =&\,\boxed{\frac{21}{16}\sin2\theta}. \end{align*}

Thus

k=2116.\boxed{k=\frac{21}{16}}.

(c)

解法一

思路

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承接 (b)，因为 0<2θ<π0<2\theta<\pi，所以 sin⁡2θ\sin2\theta 的最大值是 11，在 θ=π/4\theta=\pi/4 时取得。

答题过程

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From part (b),

Area⁡(OAP)=2116sin⁡2θ.\operatorname{Area}(OAP) =\frac{21}{16}\sin2\theta.

The maximum value of sin⁡2θ\sin2\theta is 11, attained when θ=π4\theta=\frac{\pi}{4}. Therefore, the maximum area is

2116.\boxed{\frac{21}{16}}.