Find a vector parametrization for: y^2+2x^2-2x=10

  • #1
Find a vector parametrization for: y^2+2x^2-2x=10

My attempted solution is to say that x(t)=t and y(t)= +-sqrt(-2t^2+2t+10) but I don't think it's correct to have the +- and I might need to use polar coordinates instead. I'm just not sure of the function with the extra x in it.
 

Answers and Replies

  • #2
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Well this basically looks like an off-center ellipse as you can see from the x term, so you'll want to parametrize it using the trig identity [itex]\cos^2x+\sin^2x=1[/itex]

What happens if you complete the square on the x term and make it as such?...

[tex] y=\sqrt{\frac{21}{2}}\sin{t}[/tex]

[tex]x=\sqrt{\frac{21}{4}}\left(\cos{t}+\frac{1}{2}\right)[/tex]
 

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