The curve of the intersection of the surface $x^2+y^2-2z^2=1$ and the plane $y=2z+1$ is given by $\varphi(t)=(\sqrt2\cos t,2\sin t-1,\sin t-1), t\in [0,2\pi)$.
I would like to know how to find it if I didn't know the answer. I've already tried to make projections and substitutions ($z=t$) without success.
Does anyone know a good other method to find curves of intersection of surfaces?
$\endgroup$2 Answers
$\begingroup$From the two equations we have:
$$x^2+(2z+1)^2-2z^2=1$$
$$x^2+2z^2+4z=0$$
$$x^2+2(z+1)^2=2$$
$$x=\sqrt{2}\cdot\sqrt{1-(z+1)^2}$$
Its only natural to set $(z+1)$ as $\sin(t)$ or $\cos(t)$, from which the result follows.
$\endgroup$ $\begingroup$Substituting $y=2z+1$ in the first equation you get $x^2+2z^2+4z=0$, which by completing the squares reduces to $$ \frac{x^2}{2} +(z+1)^2=1 $$ Now you can recognize the equation of an ellipse in the $xz$ plane centered in $(0,0,-1)$ and axes $\sqrt{2}$ and $1$, which can be parametrized by $(\sqrt{2}\cos t, 0, -1+\sin t)$ in the $xz$ plane. Now thinking this equation in three dimension, we are dealing with a cylinder whose axis is parallel to the $y$-axis and we have to intersect it with the plane of equation $y=2z+1$, which gives us $y=-2+2\sin t+1=2\sin t-1$ as we wanted.
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