# How do you find parametric equations for the tangent line

How do you find parametric equations for the tangent line to the curve with the given parametric equations $x=7{t}^{2}-4$ and $y=7{t}^{2}+4$ and $z=6t+5$ and $\left(3,11,11\right)$?
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Bob Huerta

x=3+14t
y=11+14t
z=11+6t
The point $\left(3,11,11\right)$ is for $t=1$, as you can see substituting it in the three equations of the curve.
Now let's search the generic vector tangent to the curve:
x'=14
y'=14
${z}^{\prime }=6$
So, for $t=1$ it is $\stackrel{\to }{v}\left(14,14,6\right)$
So, remembering that given a point $P\left({x}_{p},{y}_{p},{z}_{p}\right)$ and a direction $\stackrel{\to }{v}\left(a,b,c\right)$ the line that passes from that point with that direction is:
$x={x}_{p}+\alpha$
y=y_p+bt
z=z_p+ct
so the tangent is
x=3+14t
y=11+14t
z=11+6t
N.B. The direction (14,14,6) is the same that $\left(7,7,3\right)$ so the line could be written also:
x=3+7t
y=11+7t
z=11+3t
That's simpler!

Mary Goodson
I got it. thanks