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CIE 9709 2025 November Paper 32 Q7

A Level / CIE / P3

CIE 9709 2025 Nov Paper 32 Paper · Question 7

题目

Problem

The equation of a curve is 2y3−3x2y−x3=162y^3-3x^2y-x^3=16.

(a) Show that dydx=x2+2xy2y2−x2\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{x^2+2xy}{2y^2-x^2}.

(4)

(b) Hence find the coordinates of the points on the curve at which the normal is parallel to the yy-axis.

(4)
题目中文翻译

一条曲线的方程为 2y3−3x2y−x3=162y^3-3x^2y-x^3=16。

(a) 证明 dydx=x2+2xy2y2−x2\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{x^2+2xy}{2y^2-x^2}。

(4)

(b) 由此求曲线上法线平行于 yy 轴的点的坐标。

(4)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

展开

隐式求导时,x2yx^2y 要用乘积法则。把所有含导数的项收集在一起再约去公共系数。

答题过程

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Differentiating implicitly,

 6y2dydx−6xy −3x2dydx−3x2=0.\begin{align*} &\,6y^2\frac{\mathrm{d}y}{\mathrm{d}x}-6xy\\[4mm] &\,\hspace{2pt}-3x^2\frac{\mathrm{d}y}{\mathrm{d}x}-3x^2=0. \end{align*}

Collecting derivative terms,

(6y2−3x2)dydx=6xy+3x2.(6y^2-3x^2)\frac{\mathrm{d}y}{\mathrm{d}x} =6xy+3x^2.

Thus

dydx=6xy+3x26y2−3x2=x2+2xy2y2−x2.\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{6xy+3x^2}{6y^2-3x^2} =\boxed{\frac{x^2+2xy}{2y^2-x^2}}.

(b)

解法一

思路

展开

法线竖直意味着切线水平,因此导数的分子为零。求得点后,检查分母非零并代回曲线。

答题过程

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A normal parallel to the yy-axis corresponds to a horizontal tangent. Hence, by (a),

x2+2xy=x(x+2y)=0.x^2+2xy=x(x+2y)=0.

If x=0x=0, the curve gives 2y3=162y^3=16, so y=2y=2.

If x=−2yx=-2y, substitution gives

2y3−3(4y2)y−(−8y3)=16,2y^3-3(4y^2)y-(-8y^3)=16,

so −2y3=16-2y^3=16, giving y=−2y=-2 and x=4x=4.

The derivative denominators at these points are respectively 88 and −8-8, both nonzero. Therefore the required points are

(0,2)and(4,−2).\boxed{(0,2)\quad\text{and}\quad(4,-2)}.