﻿ 圓錐曲線證明題，給定雙曲線，求焦弦三角形最小面積(頁 1) - 高中的數學 - IV：線性代數 - Math Pro 數學補給站

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### 圓錐曲線證明題，給定雙曲線，求焦弦三角形最小面積

[size=4]設雙曲線方程式為T:x^2/a^2-y^2/b^2=1,-------------------(*1)[/size]
[size=4]焦點F1(c,0),F2(-c,0) ,其中c^2=a^2+b^2[/size]
[size=4]又設過F1(c,0)的直線方程式為L:y=m(x-c),即x=y/m +c ---------------(*2)[/size]
[size=4]且L與T的交點為A(x1,y1),B(x2,y2)[/size]
[size=4]將(*2)代入(*1) 整理得 (b^2-a^2*m^2)y^2+2mb^2*cy+b^2*c^2*m^2-a^2*b^2*m^2=0-----------(*3)[/size]
[size=4]則y1,y2滿足(*3)的解,且[/size]
[size=4]y1+y2=-2mb^2*c/(b^2-a^2*m^2)---------------------(*4)[/size]
[size=4]y1*y2=(b^2*c^2*m^2-a^2*b^2*m^2)/(b^2-a^2*m^2)[/size]
[size=4]=m^2*b^2(c^2-a^2)/(b^2-a^2*m^2)=m^2*b^4/(b^2-a^2*m^2)-------------------(*5)[/size]

[size=4]又三角形ABF2面積=(1/2)*2c|y1-y2|----------------------(*6)[/size]
[size=4][/size]
[size=4]先算|y1-y2|^2=(y1+y2)^2-4y1y2=[2mb^2*c/(b^2-a^2*m^2)]^2-4m^2*b^4/(b^2-a^2*m^2) (將(*4)(*5)代入)[/size]
[size=4]=[4m^2*b^4*c^2-4m^2*b^4(b^2-a^2*m^2)]/(b^2-a^2*m^2)^2[/size]
[size=4]=4m^2*b^4[[color=red]c^2-b^2[/color]+a^2*m^2])/(b^2-a^2*m^2)^2[/size]
[size=4]=4m^2*b^4[[color=red]a^2[/color]+a^2*m^2])/(b^2-a^2*m^2)^2[/size][size=4]
=4m^2*b^4*a^2(1+m^2)]/(b^2-a^2*m^2)^2

=| 2ab^2(1+m^2)^0.5/(a^2-b^2/m^2) |
>=| 2ab^2(1+m^2)^0.5/a^2|
>=| 2ab^2/a^2|=[color=red]2b^2/a=正焦弦長[/color][color=black]-------------(*7)[/color]
[color=blue](當m-> +或-無窮時,即直線L垂直x軸時,[/color]
[color=blue]|y1-y2|->2b^2/a )[/color]
[color=#0000ff][/color]

=(1/2)*2c|y1-y2|>=c*2b^2/a

[size=4][/size]
[size=4][/size]
[size=4][/size]

[[i] 本帖最後由 Ellipse 於 2011-4-9 10:24 AM 編輯 [/i]] [quote]原帖由 [i]Ellipse[/i] 於 2011-4-8 11:27 PM 發表 [url=https://math.pro/db/redirect.php?goto=findpost&pid=2902&ptid=1085][img]https://math.pro/db/images/common/back.gif[/img][/url]

[size=4]則|y1-y2|=|2mb^2*a(1+m^2)^0.5/(b^2-a^2*m^2)|
=| 2ab^2(1+m^2)^0.5/(a^2-b^2/m^2) |
>=| 2ab^2(1+m^2)^0.5/a^2|
>=| 2ab^2/a^2|=[color=red]2b^2/a=正焦弦長[/color][color=black]-------------(*7)[/color][/size]
[/quote]

$$\displaystyle (y_1-y_2)^2=\frac{4a^2b^4(m^4+m^2)}{a^4m^4-2a^2b^2m^2+b^4}$$

$$\displaystyle m > \frac{b}{a}$$

$$\displaystyle \frac{4a^2b^4(m^4+m^2)}{a^4m^4-2a^2b^2m^2+b^4} > \frac{4a^2b^4m^4}{a^4m^4}=\frac{4b^4}{a^2}$$

[[i] 本帖最後由 老王 於 2011-4-11 09:20 PM 編輯 [/i]] 這倒挺有趣的~~~

$$\displaystyle AF_2-AF_1=BF_2-BF_1=2a$$

$$\displaystyle s-AB=2a,s-AF_2=q,s-BF_2=p$$

$$\displaystyle \sqrt{s(s-AB)(s-AF_2)(s-BF_2)}=\sqrt{2apq(2a+p+q)}$$

$$\displaystyle AF_2-AF_1=BF_2-BF_1=2a$$