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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 $P_0(1,\sqrt{3}, 0)$?

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3 Answers 3

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 $a+\sqrt 3 b=0\implies b=0\implies x=1,y=\sqrt 3,z=c=\pm1$

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Why is it that $a^2+b^2+c^2=1$? –  Weierstraß Ramirez Oct 4 at 22:37
    
@WeierstraßRamirez, If $a^2+b^2+c^2=k^2\ne0,$ we can set $a=a'\cdot k$ etc. –  lab bhattacharjee Oct 5 at 5:59

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.

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no i could not understand sorry :( –  Yigit Oct 30 '12 at 18:52

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.

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