How to find a parametric equation for the tangent line to the curve of intersection of the cylinders? How can i find a parametric equation for the tangent line to the curve of intersection of the cylinders $x^2 + y^2 = 4$ and and $x^2 + z^2 = 1$ at the point $P_0(1,\sqrt{3}, 0)$?
 A: The equation of any line passing through $(1,\sqrt 3,0)$ can be written as
$\frac{x-1}a=\frac{y-\sqrt 3}b=\frac z c$ where $a^2+b^2+c^2=1$
So,$cx=az+c,cy=bz+\sqrt 3c$
Putting the values of $x,y$ in  $x^2+y^2=4,$
$(az+c)^2+(bz+\sqrt 3c)^2=4c^2$
$(a^2+b^2)z^2+2zc(a+\sqrt 3 b)=0$
But this is a quadratic in $z,$, each root represent the $z$ co-ordinate of the intersection.
For tangency, both root should be same, so $\{2c(a+\sqrt 3 b)\}^2=4(a^2+b^2)0$
$\implies c(a+\sqrt 3 b)=0--->(1)$
Putting the value of $x$ in $x^2+z^2=1,$
$(az+c)^2+c^2z^2=c^2\implies (a^2+c^2)z^2+2caz=0$
So like 1st case, $(2ca)^2=4(a^2+c^2)0\implies ca=0--->(2)$
Form $(1)$ and $(2)$,
if $c=0,a^2+b^2=1\implies x=a+1,y=b+\sqrt 3,z=0 $
if $c\ne 0,a=0$ and $a+\sqrt 3 b=0\implies b=0\implies x=1,y=\sqrt 3,z=c=\pm1$
A: Hint: Find the tangent planes at $P_0$ and take their intersection, that will be tangent line of the intersections.
For the planes: 1. the horizontal circle has a normal vector $(\sqrt3,-1,0)$ at $P_0$, and of course $(0,0,1)$ is always a direction of the tangent plane. Similarly you can find 2 vectors for the 2nd one.
A: take the cross product of the gradients of both functions evaluated at the point to obtain the tangent vector, assign a parameter to it, then add the point vector. 
