close
標題:
Maths-解析幾何
發問:
已知圓族F:x2+y2-4kx-8k2y+(4k2+16k4-9)=0,其中k為任意實數。 (a)証明F中每個圓的半徑皆為3。 (b)求F中圓心的軌距方程。 (c)F中有二圓C1及C2,其圓心的坐標分別為G1(a+1,y1)及G2(a-1,y2),其中a>0。若C1>C2外切。求 (i)a的值, (ii)C1與C2的方程, (iii)C1與C2在切點處的公切線方程。
最佳解答:
(a) x^2+y^2-4kx-8k^2y+(4k^2+16k^4-9)=0 (x-2k)^2+(y-4k^2)^2=9 =>r=3 (b) 令F中圓心(x,y)=(2k,4k^2) =>y=x^2 (c) 二圓C1及C2,其圓心的坐標分別為G1(a+1,y1)及G2(a-1,y2) 因外切G1G2=6 而y1=(a+1)^2,y2=(a-1)^2 [G1G2]^2=(a+1-a+1)^2+[(a+1)^2-(a-1)^2]^2 =4+16a^2 令16a^2+4=36 =>a=sqrt2 (ii) C1: (x-√2-1)^2+(y-3-2√2)^2=3 C2: (x-√2+1)^2+(y-3+2√2)^2=3 (iii) C1與C2在切點處的公切線方程 C1-C2: {(-2√2-2)x+(√2+1)^2+(-6-4√2)y+(3+2√2)^2}-{(2√2-2)x+(√2-1)^2+(6-4√2)y+(3-2√2)^2}=0 -4√2x+4√2-12y+8√2=0 4√2x+12y-12√2=0 √2x+3y-3√2=0 2009-05-01 14:02:17 補充： Can you give me the answer? 2009-05-01 15:38:40 補充： C1:[x-(√2 +1)]^2 + [y- (√2+1)^2 ]^2 =3^2 C2:[x-(√2 -1)]^2 + [y- (√2-1)^2 ]^2 =3^2 C1-C2: [2(√2 -1)-2(√2 +1)]x+[2(√2 -1)^2-2(√2 +1)^2]y+4√2+12√2 = 0 -4x-8√2y+16√2=0 x+2√2y-4√2=0
Maths-解析幾何
發問:
已知圓族F:x2+y2-4kx-8k2y+(4k2+16k4-9)=0,其中k為任意實數。 (a)証明F中每個圓的半徑皆為3。 (b)求F中圓心的軌距方程。 (c)F中有二圓C1及C2,其圓心的坐標分別為G1(a+1,y1)及G2(a-1,y2),其中a>0。若C1>C2外切。求 (i)a的值, (ii)C1與C2的方程, (iii)C1與C2在切點處的公切線方程。
最佳解答:
(a) x^2+y^2-4kx-8k^2y+(4k^2+16k^4-9)=0 (x-2k)^2+(y-4k^2)^2=9 =>r=3 (b) 令F中圓心(x,y)=(2k,4k^2) =>y=x^2 (c) 二圓C1及C2,其圓心的坐標分別為G1(a+1,y1)及G2(a-1,y2) 因外切G1G2=6 而y1=(a+1)^2,y2=(a-1)^2 [G1G2]^2=(a+1-a+1)^2+[(a+1)^2-(a-1)^2]^2 =4+16a^2 令16a^2+4=36 =>a=sqrt2 (ii) C1: (x-√2-1)^2+(y-3-2√2)^2=3 C2: (x-√2+1)^2+(y-3+2√2)^2=3 (iii) C1與C2在切點處的公切線方程 C1-C2: {(-2√2-2)x+(√2+1)^2+(-6-4√2)y+(3+2√2)^2}-{(2√2-2)x+(√2-1)^2+(6-4√2)y+(3-2√2)^2}=0 -4√2x+4√2-12y+8√2=0 4√2x+12y-12√2=0 √2x+3y-3√2=0 2009-05-01 14:02:17 補充： Can you give me the answer? 2009-05-01 15:38:40 補充： C1:[x-(√2 +1)]^2 + [y- (√2+1)^2 ]^2 =3^2 C2:[x-(√2 -1)]^2 + [y- (√2-1)^2 ]^2 =3^2 C1-C2: [2(√2 -1)-2(√2 +1)]x+[2(√2 -1)^2-2(√2 +1)^2]y+4√2+12√2 = 0 -4x-8√2y+16√2=0 x+2√2y-4√2=0
此文章來自奇摩知識+如有不便請留言告知
其他解答:
文章標籤
全站熱搜
留言列表
