Contact us
Choose a category below so we can get back to you as quickly as possible.
設中點B(x,y)連接OB,由垂徑定理可知OB⊥AB,再由畢氏定理可得OA^2+AB^2=a^2於是有x^2+y^2+(x-a)^2+y^2=a^22x^2+2y^2-2ax=0配成標準式:(x-a/2)^2+y^2=(a^2)/4其中x的取值範圍為[0,R^2/a)其實這很好理解,還是由垂徑…
We and our partners use cookies and other technologies to analyze traffic and optimize your experience. View more info and control your cookies settings at any time in our Cookies Policy.