题目
Problem
The hyperbola H has equation
25x2−4y2=1.
The line l has equation y=mx+c, where m and c are constants.
Given that l is a tangent to H,
(a) show that 25m2=4+c2.
(b) Hence find the equations of the tangents to H that pass through the point (1,2).
(c) Find the coordinates of the point of contact each of these tangents makes with H.
(12)
题目中文翻译
双曲线 H 的方程为
25x2−4y2=1
直线 l 的方程为 y=mx+c,其中 m 和 c 为常数。
已知 l 是 H 的一条切线,
(a) 证明 25m2=4+c2。
(b) 因此求经过点 (1,2) 的 H 的切线方程。
(c) 求每条切线与 H 的切点坐标。
解答
(a)
解法一
思路
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把直线 y=mx+c 代入双曲线,得到关于 x 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,c 的关系。
答题过程
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Substituting y=mx+c into the hyperbola gives
25x2−4(mx+c)2=1.
Multiplying by 100 and collecting powers of x,
(4−25m2)x2−50mcx−(25c2+100)=0.
For the line to be tangent, this quadratic must have a repeated root. Therefore its discriminant is zero:
(−50mc)2−4(4−25m2)(−25c2−100)=0.
Thus
0==2500m2c2+4(4−25m2)(25c2+100)400c2+1600−10000m2.
Dividing by 400 gives
25m2=4+c2.
解法二
思路
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官方参数法令双曲线上切点为 (5sect,2tant)。利用参数求导求切线斜率,再从点斜式读出 m、c,最后用三角恒等式验证目标关系。
答题过程
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Parametrise the hyperbola by
x=5sect,y=2tant.
Then
dxdy=5secttant2sec2t=5tant2sect.
The tangent at (5sect,2tant) is therefore
y−2tant=5tant2sect(x−5sect).
Hence
m=5tant2sect=5sint2
and
c===2tant−5msect2tant−tant2sec2t−tant2.
Thus
25m2=sin2t4,
while
4+c2=4+4cot2t=4csc2t=sin2t4.
Therefore
25m2=4+c2.
(b)
解法一
思路
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经过 (1,2) 给出 m+c=2。将 c=2−m 代入 (a) 的切线条件,解关于 m 的二次方程,再分别求截距 c。
答题过程
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Since the tangent passes through (1,2),
2=m+c,
so c=2−m. Using part (a),
25m2=4+(2−m)2.
Therefore
24m2+4m−8=6m2+m−2=(3m+2)(2m−1)=0,0,0.
Hence
m=21orm=−32.
The corresponding values of c=2−m are 3/2 and 8/3. Thus the two tangents are
y=21x+23
and
y=−32x+38.
(c)
解法一
思路
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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 x 坐标,再代回直线求 y。
答题过程
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For y=21x+23, substitution into the hyperbola gives
25x2−16(x+3)2=1.
Multiplying by 400 and simplifying,
9x2+150x+625=0,
so
(3x+25)2=0.
Thus x=−25/3, and
y=21(−325)+23=−38.
For y=−32x+38, substitution gives
25x2−9(x−4)2=1.
Multiplying by 225 and simplifying,
16x2−200x+625=0,
so
(4x−25)2=0.
Thus x=25/4, and
y=−32(425)+38=−23.
Therefore the points of contact are
(−325,−38)
and
(425,−23).