#27555:


cges30901 (cges30901)

學校 : 不指定學校
編號 : 30877
來源 : [39.9.74.255]
最後登入時間 :
2024-10-14 22:20:08
b603. 拋物線方程式 | From: [180.217.112.205] | 發表日期 : 2021-10-13 16:42

<img src="https://latex.codecogs.com/svg.image?\bg_white&space;\\(y-y_1)=k(x-x_1)^2&space;\\(y_2-y_1)=k(x_2-x_1)^2\\k=\frac{(y_2-y_1)}{(x_2-x_1)^2}\\(x_2-x_1)^2(y-y_1)=(y_2-y_1)(x-x_1)^2\\(x_2-x_1)^2y=(y_2-y_1)(x^2-2x_1x&plus;{x_1}^2)&plus;(x_2-x_1)^2y_1\\(x_2-x_1)^2y=(y_2-y_1)x^2&plus;2x_1(y_1-y_2)x&plus;(y_2-y_1){x_1}^2&plus;(x_2-x_1)^2y_1\\&space;a=(x_2-x_1)^2\\b=&space;(y_2-y_1)\\c=&space;2x_1(y_1-y_2)\\d=&space;(y_2-y_1){x_1}^2&plus;(x_2-x_1)^2y_1" title="\bg_white \\(y-y_1)=k(x-x_1)^2 \\(y_2-y_1)=k(x_2-x_1)^2\\k=\frac{(y_2-y_1)}{(x_2-x_1)^2}\\(x_2-x_1)^2(y-y_1)=(y_2-y_1)(x-x_1)^2\\(x_2-x_1)^2y=(y_2-y_1)(x^2-2x_1x+{x_1}^2)+(x_2-x_1)^2y_1\\(x_2-x_1)^2y=(y_2-y_1)x^2+2x_1(y_1-y_2)x+(y_2-y_1){x_1}^2+(x_2-x_1)^2y_1\\ a=(x_2-x_1)^2\\b= (y_2-y_1)\\c= 2x_1(y_1-y_2)\\d= (y_2-y_1){x_1}^2+(x_2-x_1)^2y_1" />

 
#27556: Re:公式


cges30901 (cges30901)

學校 : 不指定學校
編號 : 30877
來源 : [39.9.74.255]
最後登入時間 :
2024-10-14 22:20:08
b603. 拋物線方程式 | From: [180.217.112.205] | 發表日期 : 2021-10-13 16:43


https://latex.codecogs.com/svg.image?\bg_white&space;\\(y-y_1)=k(x-x_1)^2&space;\\(y_2-y_1)=k(x_2-x_1)^2\\k=\frac{(y_2-y_1)}{(x_2-x_1)^2}\\(x_2-x_1)^2(y-y_1)=(y_2-y_1)(x-x_1)^2\\(x_2-x_1)^2y=(y_2-y_1)(x^2-2x_1x&plus;{x_1}^2)&plus;(x_2-x_1)^2y_1\\(x_2-x_1)^2y=(y_2-y_1)x^2&plus;2x_1(y_1-y_2)x&plus;(y_2-y_1){x_1}^2&plus;(x_2-x_1)^2y_1\\&space;a=(x_2-x_1)^2\\b=&space;(y_2-y_1)\\c=&space;2x_1(y_1-y_2)\\d=&space;(y_2-y_1){x_1}^2&plus;(x_2-x_1)^2y_1" title="\bg_white \\(y-y_1)=k(x-x_1)^2 \\(y_2-y_1)=k(x_2-x_1)^2\\k=\frac{(y_2-y_1)}{(x_2-x_1)^2}\\(x_2-x_1)^2(y-y_1)=(y_2-y_1)(x-x_1)^2\\(x_2-x_1)^2y=(y_2-y_1)(x^2-2x_1x+{x_1}^2)+(x_2-x_1)^2y_1\\(x_2-x_1)^2y=(y_2-y_1)x^2+2x_1(y_1-y_2)x+(y_2-y_1){x_1}^2+(x_2-x_1)^2y_1\\ a=(x_2-x_1)^2\\b= (y_2-y_1)\\c= 2x_1(y_1-y_2)\\d= (y_2-y_1){x_1}^2+(x_2-x_1)^2y_1

 
#27557: Re:公式


cges30901 (cges30901)

學校 : 不指定學校
編號 : 30877
來源 : [39.9.74.255]
最後登入時間 :
2024-10-14 22:20:08
b603. 拋物線方程式 | From: [180.217.112.205] | 發表日期 : 2021-10-13 16:45


https://latex.codecogs.com/svg.image?\bg_white&space;\\(y-y_1)=k(x-x_1)^2&space;\\(y_2-y_1)=k(x_2-x_1)^2\\k=\frac{(y_2-y_1)}{(x_2-x_1)^2}\\(x_2-x_1)^2(y-y_1)=(y_2-y_1)(x-x_1)^2\\(x_2-x_1)^2y=(y_2-y_1)(x^2-2x_1x+{x_1}^2)+(x_2-x_1)^2y_1\\(x_2-x_1)^2y=(y_2-y_1)x^2+2x_1(y_1-y_2)x+(y_2-y_1){x_1}^2+(x_2-x_1)^2y_1\\&space;a=(x_2-x_1)^2\\b=&space;(y_2-y_1)\\c=&space;2x_1(y_1-y_2)\\d=&space;(y_2-y_1){x_1}^2+(x_2-x_1)^2y_1" title="\bg_white \\(y-y_1)=k(x-x_1)^2 \\(y_2-y_1)=k(x_2-x_1)^2\\k=\frac{(y_2-y_1)}{(x_2-x_1)^2}\\(x_2-x_1)^2(y-y_1)=(y_2-y_1)(x-x_1)^2\\(x_2-x_1)^2y=(y_2-y_1)(x^2-2x_1x+{x_1}^2)+(x_2-x_1)^2y_1\\(x_2-x_1)^2y=(y_2-y_1)x^2+2x_1(y_1-y_2)x+(y_2-y_1){x_1}^2+(x_2-x_1)^2y_1\\ a=(x_2-x_1)^2\\b= (y_2-y_1)\\c= 2x_1(y_1-y_2)\\d= (y_2-y_1){x_1}^2+(x_2-x_1)^2y_1

不知道要怎麼插入圖片...
[tex]\bg_white \\(y-y_1)=k(x-x_1)^2 \\(y_2-y_1)=k(x_2-x_1)^2\\k=\frac{(y_2-y_1)}{(x_2-x_1)^2}\\(x_2-x_1)^2(y-y_1)=(y_2-y_1)(x-x_1)^2\\(x_2-x_1)^2y=(y_2-y_1)(x^2-2x_1x+{x_1}^2)+(x_2-x_1)^2y_1\\(x_2-x_1)^2y=(y_2-y_1)x^2+2x_1(y_1-y_2)x+(y_2-y_1){x_1}^2+(x_2-x_1)^2y_1\\ a=(x_2-x_1)^2\\b= (y_2-y_1)\\c= 2x_1(y_1-y_2)\\d= (y_2-y_1){x_1}^2+(x_2-x_1)^2y_1[/tex]

 
#41982: Re: 公式


yizhe (yizhe)

學校 : 不指定學校
編號 : 221950
來源 : [36.228.2.49]
最後登入時間 :
2024-11-12 14:38:47
b603. 拋物線方程式 | From: [1.162.65.215] | 發表日期 : 2024-09-16 09:56


https://latex.codecogs.com/svg.image?\bg_white&space;\\(y-y_1)=k(x-x_1)^2&space;\\(y_2-y_1)=k(x_2-x_1)^2\\k=\frac{(y_2-y_1)}{(x_2-x_1)^2}\\(x_2-x_1)^2(y-y_1)=(y_2-y_1)(x-x_1)^2\\(x_2-x_1)^2y=(y_2-y_1)(x^2-2x_1x+{x_1}^2)+(x_2-x_1)^2y_1\\(x_2-x_1)^2y=(y_2-y_1)x^2+2x_1(y_1-y_2)x+(y_2-y_1){x_1}^2+(x_2-x_1)^2y_1\\&space;a=(x_2-x_1)^2\\b=&space;(y_2-y_1)\\c=&space;2x_1(y_1-y_2)\\d=&space;(y_2-y_1){x_1}^2+(x_2-x_1)^2y_1" title="\bg_white \\(y-y_1)=k(x-x_1)^2 \\(y_2-y_1)=k(x_2-x_1)^2\\k=\frac{(y_2-y_1)}{(x_2-x_1)^2}\\(x_2-x_1)^2(y-y_1)=(y_2-y_1)(x-x_1)^2\\(x_2-x_1)^2y=(y_2-y_1)(x^2-2x_1x+{x_1}^2)+(x_2-x_1)^2y_1\\(x_2-x_1)^2y=(y_2-y_1)x^2+2x_1(y_1-y_2)x+(y_2-y_1){x_1}^2+(x_2-x_1)^2y_1\\ a=(x_2-x_1)^2\\b= (y_2-y_1)\\c= 2x_1(y_1-y_2)\\d= (y_2-y_1){x_1}^2+(x_2-x_1)^2y_1

不知道要怎麼插入圖片...
[tex]\bg_white \\(y-y_1)=k(x-x_1)^2 \\(y_2-y_1)=k(x_2-x_1)^2\\k=\frac{(y_2-y_1)}{(x_2-x_1)^2}\\(x_2-x_1)^2(y-y_1)=(y_2-y_1)(x-x_1)^2\\(x_2-x_1)^2y=(y_2-y_1)(x^2-2x_1x+{x_1}^2)+(x_2-x_1)^2y_1\\(x_2-x_1)^2y=(y_2-y_1)x^2+2x_1(y_1-y_2)x+(y_2-y_1){x_1}^2+(x_2-x_1)^2y_1\\ a=(x_2-x_1)^2\\b= (y_2-y_1)\\c= 2x_1(y_1-y_2)\\d= (y_2-y_1){x_1}^2+(x_2-x_1)^2y_1[/tex]



 
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