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CIE 9709 2024 June Paper 12 Q10

A Level / CIE / P1

CIE 9709 2024 June Paper 12 Paper · Question 10

题目

Problem

The equation of a curve is y=(52x)32+5y = (5 - 2x)^{\frac{3}{2}} + 5 for x<52x < \frac{5}{2}.

(a) A point PP is moving along the curve in such a way that the yy-coordinate of point PP is decreasing at 55 units per second.

Find the rate at which the xx-coordinate of point PP is increasing when y=32y = 32.

[4]

(b) Point AA on the curve has yy-coordinate 3232. Point BB on the curve is such that the gradient of the curve at BB is 3-3.

Find the equation of the perpendicular bisector of ABAB. Give your answer in the form ax+by+c=0ax + by + c = 0, where aa, bb and cc are integers.

[6]
题目中文翻译

一条曲线的方程为 y=(52x)32+5y = (5 - 2x)^{\frac{3}{2}} + 5,其中 x<52x < \frac{5}{2}

(a) 点 PP 沿曲线运动,且点 PPyy 坐标以每秒 55 个单位的速度减小。

y=32y = 32 时,求点 PPxx 坐标增加的速率。

[4]

(b) 曲线上的点 AAyy 坐标为 3232。曲线上的点 BB 满足曲线在 BB 处的梯度为 3-3

ABAB 的垂直平分线方程。答案写成 ax+by+c=0ax + by + c = 0 的形式,其中 aabbcc 是整数。

[6]

解答