I was trying to figure out the solution for a problem then I encountered a doubt. First of all, here is the question for clarity:
Show that $2x^2+5y^2+4xy+2yz-4zx+16x+22y-10z-18=0$ is the equation of the cylinder which passes through the point $(3,-1,1)$ and has the axis $\frac{x-1}{2}=\frac{y+3}{-1}=\frac{z-2}{1}$.
My Approach:
Now, I checked, whether $(3,-1,1)$ satifies the equation of the given cylinder, and yes it satifies (first part done), now I have to prove that $\frac{x-1}{2}=\frac{y+3}{-1}=\frac{z-2}{1}$ is the equation of the axis of the given cylinder.
We know, a generator lying on the cylinder will have the same dr's as the axis (as they are both parallel lines), and we can assume that generator line to pass through $(3,-1,1)$, and hence, the equation of that generator line will be:
$\frac{x-3}{2}=\frac{y+1}{-1}=\frac{z-1}{1}=r$.
Now, any general point of the above generating line goes this way:
$x=2r+3, y=-r-1, z=r+1$.
Now, since, the generating line lies on the cylinder, each and every general point of the generating line must satisfy the equation of the cylinder. I tried substituting the general point coordinates, i.e., ($x=2r+3, y=-r-1, z=r+1$) in the given equation of the cylinder.
Then I found out:
i) coefficient of $r^2$ and $r$ is equal to 0.
ii) $LHS\neq0(RHS)$
Now, "coefficient of $r^2$ and $r$ is equal to 0" would directly mean that the generator line fully lies on the cylinder (thus increasing the probability that given axial equation is the equation of the given cylinder), but " $LHS\neq0(RHS)$ " would tell me that each and every point of the generator line is not satisfying the equation of the cylinder. Thus, because of this problem I am not able to solve this problem.
What should I do? Is my thought process/approach correct? How to prove that the given axial line equation is the equation of the given cylinder?