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anonymous
 5 years ago
How do I parametrize the curve of intersection between the surfaces z=x^2+y^2 and 2x4yz1=0?
anonymous
 5 years ago
How do I parametrize the curve of intersection between the surfaces z=x^2+y^2 and 2x4yz1=0?

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Owlfred
 5 years ago
Best ResponseYou've already chosen the best response.0Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[z=x^2+y^2\quad;\quad 2x4yz1=0\quad,\quad z=2x4y1\] \[2x4y1=x^2+y^2\] \[1=x^2+y^22x+4y=(x1)^2+(y+2)^214\] \[\Rightarrow (x1)^2+(y+2)^2=4\quad\Rightarrow x=1+2\cos{t}\,,\,y=2+2\sin{t}\] \[z=(1+2\cos{t})^2+(2+2\sin{t})^2=9+4\cos{t}8\sin{t}\,,\,t\in[0,2\pi]\]
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