题目
Problem
6. The parabola C C C y 2 = 36 x y^2 = 36x y 2 = 36 x
The point P ( 9 t 2 , 18 t ) P(9t^2, 18t) P ( 9 t 2 , 18 t ) t ≠ 0 t \neq 0 t = 0 C C C
(a) Use calculus to show that the normal to C C C P P P
y + t x = 9 t 3 + 18 t y + tx = 9t^3 + 18t y + t x = 9 t 3 + 18 t
(4)
(b) Hence find the equations of the two normals to C C C ( 54 , 0 ) (54, 0) ( 54 , 0 ) y = p x + q y = px + q y = p x + q p p p q q q
(4)
Given that
the normals found in part (b) intersect the directrix of C C C A A A B B B
the point F F F C C C
(c) determine the area of triangle A F B AFB A F B
(3)
题目中文翻译
抛物线 C C C y 2 = 36 x y^2 = 36x y 2 = 36 x
点 P ( 9 t 2 , 18 t ) P(9t^2, 18t) P ( 9 t 2 , 18 t ) t ≠ 0 t \neq 0 t = 0 C C C
(a) 利用微积分证明 C C C P P P
y + t x = 9 t 3 + 18 t y + tx = 9t^3 + 18t y + t x = 9 t 3 + 18 t
(b) 由此求经过点 ( 54 , 0 ) (54, 0) ( 54 , 0 ) C C C y = p x + q y = px + q y = p x + q p p p q q q
已知
(b) 中求得的法线与 C C C A A A B B B
点 F F F C C C
(c) 求三角形 A F B AFB A F B
解答
(a)
解法一
思路
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对抛物线方程隐函数求导，在参数点 P P P y = 18 t y=18t y = 18 t 1 / t 1/t 1/ t
答题过程
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Differentiating y 2 = 36 x y^2=36x y 2 = 36 x x x x
2 y d y d x = 36 , 2y\frac{\mathrm{d}y}{\mathrm{d}x}=36, 2 y d x d y = 36 , so
d y d x = 18 y . \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{18}{y}. d x d y = y 18 . At P ( 9 t 2 , 18 t ) P(9t^2,18t) P ( 9 t 2 , 18 t )
m T = 18 18 t = 1 t . m_T=\frac{18}{18t}=\frac1t. m T = 18 t 18 = t 1 . Since t ≠ 0 t\ne0 t = 0
m N = − 1 m T = − t . m_N=-\frac1{m_T}=-t. m N = − m T 1 = − t . The normal at P P P
y − 18 t = − t ( x − 9 t 2 ) . y-18t=-t(x-9t^2). y − 18 t = − t ( x − 9 t 2 ) . Expanding and rearranging,
y − 18 t = − t x + 9 t 3 , y + t x = 9 t 3 + 18 t . \begin{align*}
y-18t=&\,-tx+9t^3,\\
y+tx=&\,9t^3+18t.
\end{align*} y − 18 t = y + t x = − t x + 9 t 3 , 9 t 3 + 18 t . Hence the required equation is
y + t x = 9 t 3 + 18 t . \boxed{y+tx=9t^3+18t}. y + t x = 9 t 3 + 18 t .
解法二
思路
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官方评分资料也允许直接对参数方程求导。由 x = 9 t 2 x=9t^2 x = 9 t 2 y = 18 t y=18t y = 18 t d x / d t \mathrm{d}x/\mathrm{d}t d x / d t d y / d t \mathrm{d}y/\mathrm{d}t d y / d t d y / d x \mathrm{d}y/\mathrm{d}x d y / d x
答题过程
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From
x = 9 t 2 , y = 18 t , x=9t^2,
\qquad
y=18t, x = 9 t 2 , y = 18 t , we have
d x d t = 18 t , d y d t = 18. \frac{\mathrm{d}x}{\mathrm{d}t}=18t,
\qquad
\frac{\mathrm{d}y}{\mathrm{d}t}=18. d t d x = 18 t , d t d y = 18. Therefore,
d y d x = d y / d t d x / d t = 18 18 t = 1 t . \begin{align*}
\frac{\mathrm{d}y}{\mathrm{d}x}
=&\,\frac{\mathrm{d}y/\mathrm{d}t}
{\mathrm{d}x/\mathrm{d}t}\\
=&\,\frac{18}{18t}\\
=&\,\frac1t.
\end{align*} d x d y = = = d x / d t d y / d t 18 t 18 t 1 . Thus the normal gradient is − t -t − t P ( 9 t 2 , 18 t ) P(9t^2,18t) P ( 9 t 2 , 18 t )
y − 18 t = − t ( x − 9 t 2 ) , y + t x = 9 t 3 + 18 t . \begin{align*}
y-18t=&\,-t(x-9t^2),\\
y+tx=&\,\boxed{9t^3+18t}.
\end{align*} y − 18 t = y + t x = − t ( x − 9 t 2 ) , 9 t 3 + 18 t .
(b)
解法一
思路
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承接 (a)，将固定点 ( 54 , 0 ) (54,0) ( 54 , 0 ) t t t t ≠ 0 t\ne0 t = 0
答题过程
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Substituting ( x , y ) = ( 54 , 0 ) (x,y)=(54,0) ( x , y ) = ( 54 , 0 )
54 t = 9 t 3 + 18 t . 54t=9t^3+18t. 54 t = 9 t 3 + 18 t . Hence
9 t 3 − 36 t = 0 , 9 t ( t 2 − 4 ) = 0. \begin{align*}
9t^3-36t=&\,0,\\
9t(t^2-4)=&\,0.
\end{align*} 9 t 3 − 36 t = 9 t ( t 2 − 4 ) = 0 , 0. Since t ≠ 0 t\ne0 t = 0
t = 2 or t = − 2. t=2
\qquad\text{or}\qquad
t=-2. t = 2 or t = − 2. For t = 2 t=2 t = 2
y + 2 x = 9 ( 2 3 ) + 18 ( 2 ) , y = − 2 x + 108. \begin{align*}
y+2x=&\,9(2^3)+18(2),\\
y=&\,-2x+108.
\end{align*} y + 2 x = y = 9 ( 2 3 ) + 18 ( 2 ) , − 2 x + 108. For t = − 2 t=-2 t = − 2
y − 2 x = 9 ( − 2 ) 3 + 18 ( − 2 ) , y = 2 x − 108. \begin{align*}
y-2x=&\,9(-2)^3+18(-2),\\
y=&\,2x-108.
\end{align*} y − 2 x = y = 9 ( − 2 ) 3 + 18 ( − 2 ) , 2 x − 108. Therefore, the two normals are
y = − 2 x + 108 \boxed{y=-2x+108} y = − 2 x + 108 and
y = 2 x − 108 . \boxed{y=2x-108}. y = 2 x − 108 .
(c)
解法一
思路
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由 y 2 = 4 a x y^2=4ax y 2 = 4 a x A , B A,B A , B A B AB A B
答题过程
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Comparing y 2 = 36 x y^2=36x y 2 = 36 x y 2 = 4 a x y^2=4ax y 2 = 4 a x a = 9 a=9 a = 9
x = − 9 , x=-9, x = − 9 , and the focus is
F = ( 9 , 0 ) . F=(9,0). F = ( 9 , 0 ) . At x = − 9 x=-9 x = − 9
y = − 2 ( − 9 ) + 108 = 126 y=-2(-9)+108=126 y = − 2 ( − 9 ) + 108 = 126 and
y = 2 ( − 9 ) − 108 = − 126. y=2(-9)-108=-126. y = 2 ( − 9 ) − 108 = − 126. Thus the two intersections are
A = ( − 9 , 126 ) , B = ( − 9 , − 126 ) . \begin{align*}
A=&\,(-9,126),\\
B=&\,(-9,-126).
\end{align*} A = B = ( − 9 , 126 ) , ( − 9 , − 126 ) . Therefore,
A B = 126 − ( − 126 ) = 252 , AB=126-(-126)=252, A B = 126 − ( − 126 ) = 252 , and the perpendicular distance from F F F
9 − ( − 9 ) = 18. 9-(-9)=18. 9 − ( − 9 ) = 18. Hence
Area of △ A F B = 1 2 × 252 × 18 = 2268 square units . \begin{align*}
\text{Area of }\triangle AFB
=&\,\frac12\times252\times18\\
=&\,\boxed{2268\text{ square units}}.
\end{align*} Area of △ A F B = = 2 1 × 252 × 18 2268 square units .
解法二
思路
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官方替代路线使用鞋带公式。先沿用解法一所得的三个顶点坐标，再按同一方向排列并计算有向面积，最后取绝对值。
答题过程
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Using
A = ( − 9 , 126 ) , F = ( 9 , 0 ) , B = ( − 9 , − 126 ) . \begin{align*}
A=&\,(-9,126),\\
F=&\,(9,0),\\
B=&\,(-9,-126).
\end{align*} A = F = B = ( − 9 , 126 ) , ( 9 , 0 ) , ( − 9 , − 126 ) . the shoelace formula gives
Area = 1 2 ∣ ( − 9 ) ( 0 ) + 9 ( − 126 ) + ( − 9 ) ( 126 ) − ( 126 ( 9 ) + 0 ( − 9 ) + ( − 126 ) ( − 9 ) ) ∣ = 1 2 ∣ − 4536 ∣ = 2268 square units . \begin{align*}
\text{Area}
=&\,\frac12\Big|
(-9)(0)+9(-126)\\
&\,\hspace{12pt}+(-9)(126)\\
&\,\hspace{14pt}-\big(126(9)+0(-9)\\
&\,\hspace{28pt}+(-126)(-9)\big)
\Big|\\
=&\,\frac12|-4536|\\
=&\,\boxed{2268\text{ square units}}.
\end{align*} Area = = = 2 1 ( − 9 ) ( 0 ) + 9 ( − 126 ) + ( − 9 ) ( 126 ) − ( 126 ( 9 ) + 0 ( − 9 ) + ( − 126 ) ( − 9 ) ) 2 1 ∣ − 4536∣ 2268 square units . 