Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9709 2023 June Paper 32 Q7

A Level / CIE / P3

CIE 9709 2023 June Paper 32 Paper · Question 7

题目

Problem

The equation of a curve is 3x2+4xy+3y2=53x^2+4xy+3y^2=5.

(a) Show that dydx=−3x+2y2x+3y\dfrac{\mathrm{d}y}{\mathrm{d}x}=-\dfrac{3x+2y}{2x+3y}.

(4)

(b) Hence find the exact coordinates of the two points on the curve at which the tangent is parallel to y+2x=0y+2x=0.

(5)
题目中文翻译

一条曲线的方程为 3x2+4xy+3y2=53x^2+4xy+3y^2=5。

(a) 证明 dydx=−3x+2y2x+3y\dfrac{\mathrm{d}y}{\mathrm{d}x}=-\dfrac{3x+2y}{2x+3y}。

(4)

(b) 由此求曲线上切线平行于 y+2x=0y+2x=0 的两个点的精确坐标。

(5)
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

对 4xy4xy 使用乘积法则,其余含 yy 的项使用链式法则,再收集导数项。

答题过程

展开

Differentiating implicitly,

6x+4y+(4x+6y)dydx=0.6x+4y+(4x+6y)\frac{\mathrm{d}y}{\mathrm{d}x}=0.

Thus

(4x+6y)dydx=−(6x+4y).(4x+6y)\frac{\mathrm{d}y}{\mathrm{d}x}=-(6x+4y).

Dividing and cancelling the factor 22 gives

dydx=−3x+2y2x+3y.\boxed{\frac{\mathrm{d}y}{\mathrm{d}x} =-\frac{3x+2y}{2x+3y}}.

(b)

解法一

思路

展开

给定直线斜率为 −2-2。令(a)导数等于 −2-2,求出 x,yx,y 的线性关系,再与曲线联立;坐标的正负号须成对。

答题过程

展开

The line y+2x=0y+2x=0 has gradient −2-2. Therefore

−3x+2y2x+3y=−2.-\frac{3x+2y}{2x+3y}=-2.

Hence

3x+2y= 4x+6y,x= −4y.\begin{align*} 3x+2y=&\,4x+6y,\\[4mm] x=&\,-4y. \end{align*}

Substituting into 3x2+4xy+3y2=53x^2+4xy+3y^2=5,

48y2−16y2+3y2= 5,35y2= 5.\begin{align*} 48y^2-16y^2+3y^2=&\,5,\\[4mm] 35y^2=&\,5. \end{align*}

Thus y=±1/7y=\pm1/\sqrt7 and x=−4yx=-4y. The denominator 2x+3y=−5y2x+3y=-5y is non-zero at both points. Therefore the coordinates are

 (−47,17) and (47,−17).\begin{align*} &\,\boxed{\bigg(-\frac4{\sqrt7},\frac1{\sqrt7}\bigg)}\\[4mm] &\,\text{and }\boxed{\bigg(\frac4{\sqrt7},-\frac1{\sqrt7}\bigg)}. \end{align*}