题目
Problem
8.
The hyperbola H H H
4 x 2 − y 2 = 4. 4x^2 - y^2 = 4. 4 x 2 − y 2 = 4. (a) Write down the equations of the asymptotes of H H H
(1)
(b) Find the coordinates of the foci of H H H
(2)
The point P ( sec θ , 2 tan θ ) P(\sec\theta, 2\tan\theta) P ( sec θ , 2 tan θ ) H H H
(c) Using calculus, show that the equation of the tangent to H H H P P P
y tan θ = 2 x sec θ − 2. y\tan\theta = 2x\sec\theta - 2. y tan θ = 2 x sec θ − 2. (4)
The point V ( − 1 , 0 ) V(-1, 0) V ( − 1 , 0 ) W ( 1 , 0 ) W(1, 0) W ( 1 , 0 ) H H H Q ( sec θ , − 2 tan θ ) Q(\sec\theta, -2\tan\theta) Q ( sec θ , − 2 tan θ ) H H H P P P Q Q Q V V V W W W H H H V P VP V P W Q WQ W Q S S S
(d) show that, as θ \theta θ S S S
x 2 a 2 + y 2 b 2 = 1 \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 a 2 x 2 + b 2 y 2 = 1 where a a a b b b
(7)
题目中文翻译
双曲线 H H H
4 x 2 − y 2 = 4 4x^2 - y^2 = 4 4 x 2 − y 2 = 4 (a) 写出 H H H
(1)
(b) 求 H H H
(2)
点 P ( sec θ , 2 tan θ ) P(\sec\theta, 2\tan\theta) P ( sec θ , 2 tan θ ) H H H
(c) 利用微积分证明，H H H P P P
y tan θ = 2 x sec θ − 2 y\tan\theta = 2x\sec\theta - 2 y tan θ = 2 x sec θ − 2 (4)
点 V ( − 1 , 0 ) V(-1, 0) V ( − 1 , 0 ) W ( 1 , 0 ) W(1, 0) W ( 1 , 0 ) H H H Q ( sec θ , − 2 tan θ ) Q(\sec\theta, -2\tan\theta) Q ( sec θ , − 2 tan θ ) H H H P P P Q Q Q V V V W W W H H H V P VP V P W Q WQ W Q S S S
(d) 证明当 θ \theta θ S S S
x 2 a 2 + y 2 b 2 = 1 \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 a 2 x 2 + b 2 y 2 = 1 其中 a a a b b b
(7)
解答
(a)
解法一
思路
展开
把双曲线写成标准形式 x 2 − y 2 / 4 = 1 x^2-y^2/4=1 x 2 − y 2 /4 = 1 x 2 / a 2 − y 2 / b 2 = 1 x^2/a^2-y^2/b^2=1 x 2 / a 2 − y 2 / b 2 = 1 y = ± ( b / a ) x y=\pm(b/a)x y = ± ( b / a ) x
答题过程
展开
The equation can be written as
x 2 − y 2 4 = 1. x^2-\frac{y^2}{4}=1. x 2 − 4 y 2 = 1. Here a = 1 a=1 a = 1 b = 2 b=2 b = 2
y = 2 x and y = − 2 x . \boxed{y=2x\quad\text{and}\quad y=-2x}. y = 2 x and y = − 2 x .
(b)
解法一
思路
展开
横轴双曲线的焦距满足 c 2 = a 2 + b 2 c^2=a^2+b^2 c 2 = a 2 + b 2 a = 1 , b = 2 a=1,b=2 a = 1 , b = 2 c c c x x x
答题过程
展开
For the hyperbola,
c 2 = a 2 + b 2 = 1 2 + 2 2 = 5. c^2=a^2+b^2=1^2+2^2=5. c 2 = a 2 + b 2 = 1 2 + 2 2 = 5. Hence the foci are
( 5 , 0 ) and ( − 5 , 0 ) . \boxed{(\sqrt5,0)\quad\text{and}\quad(-\sqrt5,0)}. ( 5 , 0 ) and ( − 5 , 0 ) .
(c)
解法一
思路
展开
题目要求使用微积分，因此对双曲线隐式求导，在 P ( sec θ , 2 tan θ ) P(\sec\theta,2\tan\theta) P ( sec θ , 2 tan θ ) sec 2 θ − tan 2 θ = 1 \sec^2\theta-\tan^2\theta=1 sec 2 θ − tan 2 θ = 1
答题过程
展开
Differentiating 4 x 2 − y 2 = 4 4x^2-y^2=4 4 x 2 − y 2 = 4
8 x − 2 y d y d x = 0. 8x-2y\frac{\mathrm{d}y}{\mathrm{d}x}=0. 8 x − 2 y d x d y = 0. Therefore
d y d x = 4 x y . \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{4x}{y}. d x d y = y 4 x . At P ( sec θ , 2 tan θ ) P(\sec\theta,2\tan\theta) P ( sec θ , 2 tan θ )
4 sec θ 2 tan θ = 2 sec θ tan θ . \frac{4\sec\theta}{2\tan\theta}
=\frac{2\sec\theta}{\tan\theta}. 2 tan θ 4 sec θ = tan θ 2 sec θ . Thus the tangent is
y − 2 tan θ = 2 sec θ tan θ ( x − sec θ ) . y-2\tan\theta
=\frac{2\sec\theta}{\tan\theta}
(x-\sec\theta). y − 2 tan θ = tan θ 2 sec θ ( x − sec θ ) . Multiplying by tan θ \tan\theta tan θ
y tan θ − 2 tan 2 θ = 2 x sec θ − 2 sec 2 θ . y\tan\theta-2\tan^2\theta
=2x\sec\theta-2\sec^2\theta. y tan θ − 2 tan 2 θ = 2 x sec θ − 2 sec 2 θ . Since sec 2 θ − tan 2 θ = 1 \sec^2\theta-\tan^2\theta=1 sec 2 θ − tan 2 θ = 1
y tan θ = 2 x sec θ − 2 . \boxed{y\tan\theta=2x\sec\theta-2}. y tan θ = 2 x sec θ − 2 .
解法二
思路
展开
把 x = sec θ x=\sec\theta x = sec θ y = 2 tan θ y=2\tan\theta y = 2 tan θ θ \theta θ
答题过程
展开
Using the parametric equations,
d x d θ = sec θ tan θ , d y d θ = 2 sec 2 θ . \frac{\mathrm{d}x}{\mathrm{d}\theta}=\sec\theta\tan\theta,
\qquad
\frac{\mathrm{d}y}{\mathrm{d}\theta}=2\sec^2\theta. d θ d x = sec θ tan θ , d θ d y = 2 sec 2 θ . Hence the gradient at P P P
d y d x = 2 sec 2 θ sec θ tan θ = 2 sec θ tan θ . \frac{\mathrm{d}y}{\mathrm{d}x}
=\frac{2\sec^2\theta}{\sec\theta\tan\theta}
=\frac{2\sec\theta}{\tan\theta}. d x d y = sec θ tan θ 2 sec 2 θ = tan θ 2 sec θ . The tangent through P ( sec θ , 2 tan θ ) P(\sec\theta,2\tan\theta) P ( sec θ , 2 tan θ )
y − 2 tan θ = 2 sec θ tan θ ( x − sec θ ) . y-2\tan\theta=\frac{2\sec\theta}{\tan\theta}(x-\sec\theta). y − 2 tan θ = tan θ 2 sec θ ( x − sec θ ) . Multiplying by tan θ \tan\theta tan θ
y tan θ = 2 x sec θ − 2 sec 2 θ + 2 tan 2 θ = 2 x sec θ − 2 ( sec 2 θ − tan 2 θ ) = 2 x sec θ − 2. \begin{align*}
y\tan\theta=&\,2x\sec\theta-2\sec^2\theta+2\tan^2\theta\\
=&\,2x\sec\theta-2(\sec^2\theta-\tan^2\theta)\\
=&\,2x\sec\theta-2.
\end{align*} y tan θ = = = 2 x sec θ − 2 sec 2 θ + 2 tan 2 θ 2 x sec θ − 2 ( sec 2 θ − tan 2 θ ) 2 x sec θ − 2. Thus the tangent is y tan θ = 2 x sec θ − 2 \boxed{y\tan\theta=2x\sec\theta-2} y tan θ = 2 x sec θ − 2
(d)
解法一
思路
展开
分别由两点式写出 V P VP V P W Q WQ W Q S S S x = cos θ x=\cos\theta x = cos θ y = 2 sin θ y=2\sin\theta y = 2 sin θ
答题过程
展开
The line through V ( − 1 , 0 ) V(-1,0) V ( − 1 , 0 ) P ( sec θ , 2 tan θ ) P(\sec\theta,2\tan\theta) P ( sec θ , 2 tan θ )
V P : y = 2 tan θ sec θ + 1 ( x + 1 ) . VP:\quad
y=\frac{2\tan\theta}{\sec\theta+1}(x+1). V P : y = sec θ + 1 2 tan θ ( x + 1 ) . The line through W ( 1 , 0 ) W(1,0) W ( 1 , 0 ) Q ( sec θ , − 2 tan θ ) Q(\sec\theta,-2\tan\theta) Q ( sec θ , − 2 tan θ )
W Q : y = − 2 tan θ sec θ − 1 ( x − 1 ) . WQ:\quad
y=-\frac{2\tan\theta}{\sec\theta-1}(x-1). W Q : y = − sec θ − 1 2 tan θ ( x − 1 ) . Since P , Q , V , W P,Q,V,W P , Q , V , W tan θ ≠ 0 \tan\theta\ne0 tan θ = 0 sec θ ≠ ± 1 \sec\theta\ne\pm1 sec θ = ± 1 S S S 2 tan θ 2\tan\theta 2 tan θ
x + 1 sec θ + 1 = − x − 1 sec θ − 1 . \frac{x+1}{\sec\theta+1}
=-\frac{x-1}{\sec\theta-1}. sec θ + 1 x + 1 = − sec θ − 1 x − 1 . Therefore
( x + 1 ) ( sec θ − 1 ) = − ( x − 1 ) ( sec θ + 1 ) , (x+1)(\sec\theta-1)
=-(x-1)(\sec\theta+1), ( x + 1 ) ( sec θ − 1 ) = − ( x − 1 ) ( sec θ + 1 ) , Expanding and collecting terms,
x sec θ − x + sec θ − 1 = − x sec θ − x + sec θ + 1 , 2 x sec θ = 2 , x sec θ = 1. \begin{align*}
x\sec\theta-x+\sec\theta-1
=&\,-x\sec\theta-x+\sec\theta+1,\\
2x\sec\theta=&\,2,\\
x\sec\theta=&\,1.
\end{align*} x sec θ − x + sec θ − 1 = 2 x sec θ = x sec θ = − x sec θ − x + sec θ + 1 , 2 , 1. Hence
x = cos θ . x=\cos\theta. x = cos θ . Substituting into the equation of V P VP V P
y = 2 tan θ sec θ + 1 ( cos θ + 1 ) = 2 tan θ cos θ = 2 sin θ . \begin{align*}
y
=&\,\frac{2\tan\theta}{\sec\theta+1}
(\cos\theta+1)\\[4mm]
=&\,2\tan\theta\cos\theta\\[4mm]
=&\,2\sin\theta.
\end{align*} y = = = sec θ + 1 2 tan θ ( cos θ + 1 ) 2 tan θ cos θ 2 sin θ . Thus the locus of S S S
x 2 + y 2 4 = cos 2 θ + sin 2 θ = 1. x^2+\frac{y^2}{4}
=\cos^2\theta+\sin^2\theta
=1. x 2 + 4 y 2 = cos 2 θ + sin 2 θ = 1. Therefore S S S
x 2 1 2 + y 2 2 2 = 1 , \boxed{\frac{x^2}{1^2}+\frac{y^2}{2^2}=1}, 1 2 x 2 + 2 2 y 2 = 1 , so
a = 1 , b = 2 . \boxed{a=1,\qquad b=2}. a = 1 , b = 2 . 