How do I parametrize the curve of intersection between the surfaces z=x^2+y^2 and 2x-4y-z-1=0?

Question

owlfred · Answer

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nikvist · Answer

$$z=x^2+y^2\quad;\quad 2x-4y-z-1=0\quad,\quad z=2x-4y-1$$
$$2x-4y-1=x^2+y^2$$
$$-1=x^2+y^2-2x+4y=(x-1)^2+(y+2)^2-1-4$$
$$\Rightarrow (x-1)^2+(y+2)^2=4\quad\Rightarrow x=1+2\cos{t}\,,\,y=-2+2\sin{t}$$
$$z=(1+2\cos{t})^2+(-2+2\sin{t})^2=9+4\cos{t}-8\sin{t}\,,\,t\in[0,2\pi]$$

anonymous · Answer

Thank You!