Skip to content
CalcGospel 國際數學圖譜
返回

IAL 2020 Oct FP3 Q5

A Level / Edexcel / FP3

IAL 2020 Oct Paper · Question 5

题目

Problem

The hyperbola HH has equation

x225y24=1.\frac{x^2}{25} - \frac{y^2}{4} = 1.

The line ll has equation y=mx+cy = mx + c, where mm and cc are constants.

Given that ll is a tangent to HH,

(a) show that 25m2=4+c225m^2 = 4 + c^2.

(b) Hence find the equations of the tangents to HH that pass through the point (1,2)(1, 2).

(c) Find the coordinates of the point of contact each of these tangents makes with HH.

(12)
题目中文翻译

双曲线 HH 的方程为

x225y24=1\frac{x^2}{25} - \frac{y^2}{4} = 1

直线 ll 的方程为 y=mx+cy = mx + c,其中 mmcc 为常数。

已知 llHH 的一条切线,

(a) 证明 25m2=4+c225m^2 = 4 + c^2

(b) 因此求经过点 (1,2)(1, 2)HH 的切线方程。

(c) 求每条切线与 HH 的切点坐标。

解答

(a)

解法一

思路

展开

把直线 y=mx+cy=mx+c 代入双曲线,得到关于 xx 的二次方程。直线为切线意味着只有一个交点,因此判别式必须为零;整理即可得到 m,cm,c 的关系。

答题过程

展开

Substituting y=mx+cy=mx+c into the hyperbola gives

x225(mx+c)24=1.\frac{x^2}{25}-\frac{(mx+c)^2}{4}=1.

Multiplying by 100 and collecting powers of xx,

(425m2)x250mcx(25c2+100)=0.(4-25m^2)x^2-50mcx-(25c^2+100)=0.

For the line to be tangent, this quadratic must have a repeated root. Therefore its discriminant is zero:

(50mc)24(425m2)(25c2100)=0.(-50mc)^2 -4(4-25m^2)(-25c^2-100)=0.

Thus

0=2500m2c2+4(425m2)(25c2+100)=400c2+160010000m2.\begin{align*} 0 =&\,2500m^2c^2 +4(4-25m^2)(25c^2+100)\\[4mm] =&\,400c^2+1600-10000m^2. \end{align*}

Dividing by 400 gives

25m2=4+c2.\boxed{25m^2=4+c^2}.

解法二

思路

展开

官方参数法令双曲线上切点为 (5sect,2tant)(5\sec t,2\tan t)。利用参数求导求切线斜率,再从点斜式读出 mmcc,最后用三角恒等式验证目标关系。

答题过程

展开

Parametrise the hyperbola by

x=5sect,y=2tant.x=5\sec t, \qquad y=2\tan t.

Then

dydx=2sec2t5secttant=2sect5tant.\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{2\sec^2t}{5\sec t\tan t} =\frac{2\sec t}{5\tan t}.

The tangent at (5sect,2tant)(5\sec t,2\tan t) is therefore

y2tant=2sect5tant(x5sect).y-2\tan t =\frac{2\sec t}{5\tan t}(x-5\sec t).

Hence

m=2sect5tant=25sintm=\frac{2\sec t}{5\tan t} =\frac{2}{5\sin t}

and

c=2tant5msect=2tant2sec2ttant=2tant.\begin{align*} c =&\,2\tan t-5m\sec t\\[4mm] =&\,2\tan t-\frac{2\sec^2t}{\tan t}\\[4mm] =&\,-\frac{2}{\tan t}. \end{align*}

Thus

25m2=4sin2t,25m^2=\frac{4}{\sin^2t},

while

4+c2=4+4cot2t=4csc2t=4sin2t.4+c^2 =4+4\cot^2t =4\csc^2t =\frac{4}{\sin^2t}.

Therefore

25m2=4+c2.\boxed{25m^2=4+c^2}.

(b)

解法一

思路

展开

经过 (1,2)(1,2) 给出 m+c=2m+c=2。将 c=2mc=2-m 代入 (a) 的切线条件,解关于 mm 的二次方程,再分别求截距 cc

答题过程

展开

Since the tangent passes through (1,2)(1,2),

2=m+c,2=m+c,

so c=2mc=2-m. Using part (a),

25m2=4+(2m)2.25m^2=4+(2-m)^2.

Therefore

24m2+4m8=0,6m2+m2=0,(3m+2)(2m1)=0.\begin{align*} 24m^2+4m-8=&\,0,\\[4mm] 6m^2+m-2=&\,0,\\[4mm] (3m+2)(2m-1)=&\,0. \end{align*}

Hence

m=12orm=23.m=\frac12 \qquad\text{or}\qquad m=-\frac23.

The corresponding values of c=2mc=2-m are 3/23/2 and 8/38/3. Thus the two tangents are

y=12x+32\boxed{y=\frac12x+\frac32}

and

y=23x+83.\boxed{y=-\frac23x+\frac83}.

(c)

解法一

思路

展开

把 (b) 的每条切线分别代回双曲线。因为它们是切线,所得二次方程应为完全平方并有重根;重根就是切点的 xx 坐标,再代回直线求 yy

答题过程

展开

For y=12x+32y=\frac12x+\frac32, substitution into the hyperbola gives

x225(x+3)216=1.\frac{x^2}{25} -\frac{(x+3)^2}{16}=1.

Multiplying by 400 and simplifying,

9x2+150x+625=0,9x^2+150x+625=0,

so

(3x+25)2=0.(3x+25)^2=0.

Thus x=25/3x=-25/3, and

y=12(253)+32=83.y=\frac12\left(-\frac{25}{3}\right)+\frac32 =-\frac83.

For y=23x+83y=-\frac23x+\frac83, substitution gives

x225(x4)29=1.\frac{x^2}{25}-\frac{(x-4)^2}{9}=1.

Multiplying by 225 and simplifying,

16x2200x+625=0,16x^2-200x+625=0,

so

(4x25)2=0.(4x-25)^2=0.

Thus x=25/4x=25/4, and

y=23(254)+83=32.y=-\frac23\left(\frac{25}{4}\right)+\frac83 =-\frac32.

Therefore the points of contact are

(253,83)\boxed{\left(-\frac{25}{3},-\frac83\right)}

and

(254,32).\boxed{\left(\frac{25}{4},-\frac32\right)}.