moduły liczb zespolonych

[adrianst]

\(\displaystyle{ \left| z+i\right|+ ft| z-i\right|=8}\)

[nitka_angel]

\(\displaystyle{ \left|z+i \right| + ft| z-1\right| = 8}\)
\(\displaystyle{ \left| x+iy+i\right| + ft| x+iy - i\right| = 8}\)
\(\displaystyle{ \sqrt{x ^{2} +(y+1) ^{2} } + \sqrt{x ^{2} +(y-1) ^{2} }=8}\)
\(\displaystyle{ \sqrt{x ^{2} +(y+1) ^{2} } = 8- \sqrt{x ^{2} +(y-1) ^{2} }}\)
\(\displaystyle{ x ^{2} +y ^{2} + 2y +1= 64 - 16 \sqrt{x ^{2} +(y-1) ^{2} } +x ^{2} + y ^{2} -2y+1}\)
\(\displaystyle{ 16 \sqrt{x ^{2} +(y-1) ^{2} }= 16 -y}\)
\(\displaystyle{ 16(x ^{2} + y ^{2} -2y +1)= 256 -32y+ y ^{2}}\)
\(\displaystyle{ 16x ^{2} +16y ^{2} - 32y+16= 256 -32y +y ^{2}}\)
\(\displaystyle{ 16 x ^{2} +15 y ^{2} = 240 / : 240}\)
\(\displaystyle{ \frac{16x ^{2} }{240} + \frac{15y ^{2} }{240}=1}\)
\(\displaystyle{ \frac{x ^{2} }{15} + \frac{y ^{2} }{16} = 1}\)

więc jest to elipsa gdzie:
\(\displaystyle{ a= \sqrt{15}, b= 4}\)



Kolega z WAT-u??

[adrianst]

dzieki bardzo,

[ Dodano: 24 Października 2008, 13:57 ]
a po czym tak wnioskujesz??