Skip to content
CalcGospel 國際數學圖譜
返回

IAL 2019 May FP1 Q9

A Level / Edexcel / FP1

IAL 2019 May Paper · Question 9

Question

Problem

The parabola CC has cartesian equation y2=4axy^2 = 4ax, where aa is a positive constant.

The point P(ap2,2ap)P(ap^2, 2ap) lies on CC.

The line ll is the normal to CC at the point PP.

The line ll passes through the point BB with coordinates (10a,0)(10a, 0).

Given that p>0p > 0

(a) use calculus to find, in terms of aa only, the coordinates of PP.

The point SS is the focus of the parabola CC.

(b) Find, in terms of aa, the exact area of triangle SBPSBP.

A circle has equation

(x10a)2+y2=9a24(x - 10a)^2 + y^2 = \frac{9a^2}{4}

Given that the line ll cuts this circle at the point RR, where y>0y > 0

(c) find, in terms of aa, the distance PRPR.

中文翻译

抛物线 CC 的直角坐标方程为 y2=4axy^2 = 4ax,其中 aa 是一个正常数。

P(ap2,2ap)P(ap^2, 2ap)CC 上。

直线 llCC 在点 PP 处的法线。

直线 ll 经过坐标为 (10a,0)(10a, 0) 的点 BB

已知 p>0p > 0

(a) 用微积分方法,仅用 aa 表示,求点 PP 的坐标。

SS 是抛物线 CC 的焦点。

(b) 用 aa 表示,求三角形 SBPSBP 的精确面积。

一个圆的方程为

(x10a)2+y2=9a24(x - 10a)^2 + y^2 = \frac{9a^2}{4}

已知直线 ll 与该圆相交于点 RR,其中 y>0y > 0

(c) 用 aa 表示,求距离 PRPR

解答

(a)

解法一

思路

展开

对抛物线方程隐式求导,得到点 PP 处的切线斜率,再取负倒数求法线斜率。法线同时经过 PPBB,利用点斜式求出参数 pp;最后代回 PP 的参数坐标。

答题过程

展开

Differentiating y2=4axy^2=4ax implicitly gives

2ydydx=4a,2y\frac{\mathrm{d}y}{\mathrm{d}x}=4a,

so

dydx=2ay.\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{2a}{y}.

At P(ap2,2ap)P(ap^2,2ap), the gradient of the tangent is

2a2ap=1p.\frac{2a}{2ap}=\frac{1}{p}.

Therefore, the gradient of the normal is p-p. Since the normal passes through PP and B(10a,0)B(10a,0),

02ap=p(10aap2).0-2ap=-p(10a-ap^2).

As p>0p>0, we may divide by p-p, giving

2a=10aap2.2a=10a-ap^2.

Since a>0a>0,

p2=8.p^2=8.

The condition p>0p>0 gives

p=22.p=2\sqrt{2}.

Hence

xP=ap2=8a,yP=2ap=42a.\begin{align*} x_P =&\,ap^2=8a, \\ y_P =&\,2ap=4\sqrt{2}\,a. \end{align*}

Therefore,

P=(8a,42a).\boxed{P=(8a,4\sqrt{2}\,a)}.

(b)

解法一

思路

展开

抛物线 y2=4axy^2=4ax 的焦点是 S=(a,0)S=(a,0)。线段 SBSB 位于 xx 轴上,可作为三角形的底;点 PP 的纵坐标就是相应的高。

答题过程

展开

The focus of CC is

S=(a,0).S=(a,0).

Since B=(10a,0)B=(10a,0),

SB=10aa=9a.SB=10a-a=9a.

Using the yy-coordinate of PP found in part (a),

Area of SBP=12(9a)(42a)=182a2.\begin{align*} \text{Area of }\triangle SBP =&\,\frac{1}{2}(9a)(4\sqrt{2}\,a) \\ =&\,\boxed{18\sqrt{2}\,a^2}. \end{align*}

(c)

解法一

思路

展开

圆心是 B=(10a,0)B=(10a,0),而法线 ll 经过 BB,所以 BRBR 是圆的半径。由 yR>0y_R>0 可知 RRPP 位于从 BB 出发的同一条射线上,因此 PR=PBBRPR=PB-BR

答题过程

展开

Using P=(8a,42a)P=(8a,4\sqrt{2}\,a) and B=(10a,0)B=(10a,0),

PB=(10a8a)2+(42a)2=4a2+32a2=6a.\begin{align*} PB =&\,\sqrt{(10a-8a)^2+(4\sqrt{2}\,a)^2} \\ =&\,\sqrt{4a^2+32a^2} \\ =&\,6a. \end{align*}

The circle has centre BB and radius

BR=3a2.BR=\frac{3a}{2}.

The condition yR>0y_R>0 places RR on the ray from BB towards PP. Hence

PR=PBBR=6a3a2=9a2.\begin{align*} PR =&\,PB-BR \\ =&\,6a-\frac{3a}{2} \\ =&\,\boxed{\frac{9a}{2}}. \end{align*}