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Find the curve of intersection between the surfaces described by $-3x^2 + 5z =1$ and $4x^2 + 2y^2 = 11$. the parametrization must be so that $y = ksin(t)$ where $k $ is a positive constant, and such that the other surface is traversed counter clockwise.

Im having a hard time finding the constant $k$, and also making sense of the "traversed counter clockwise" part. I also got that $$ x = \pm \sqrt{11-2k^2sin^2(t) \over 4}$$ and i don't know what to do with the $\pm$.

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  • $\begingroup$ perhaps $2k^2=11$? $\endgroup$ – imranfat Jan 23 '18 at 20:14
  • $\begingroup$ What made you think of that? $\endgroup$ – Pame Jan 23 '18 at 20:59
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Try this:$$x=\frac{\sqrt {11}}{2}\cos t\\y=\sqrt\frac{11}{2}\sin t\\z=\frac{1+\frac{33}{4}\cos^2 t}{5}$$

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  • $\begingroup$ Thank you for any comment on this! $\endgroup$ – Mostafa Ayaz Jan 23 '18 at 20:19
  • $\begingroup$ How did you come up with them? What happened to the $\pm$ ? $\endgroup$ – Pame Jan 23 '18 at 20:32
  • $\begingroup$ Just easily substitute $y=k\sin t$ in $4x^2+2y^2=11$ and extract $x$ then substitute it in $z=1+3x^2$ $\endgroup$ – Mostafa Ayaz Jan 23 '18 at 20:33
  • $\begingroup$ I did that and got $$ x = \pm \sqrt{11-2k^2sin^2(t) \over 4}$$ I don't see how you got the expression you did. $\endgroup$ – Pame Jan 23 '18 at 20:57
  • $\begingroup$ The questions meant that you must replace $k$ with some proper value. The expression you attained is true. You only need to replace $k$ with $\sqrt\frac{11}{2}$ $\endgroup$ – Mostafa Ayaz Jan 23 '18 at 21:01

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