Answer:
Step-by-step explanation:
We have a curve (an ellipse) written as the system of equations
.
And we want to calculate the tangent at the point (3,4,9).
The idea in this problem is to consider two variables as functions of the third. Usually we consider 
and 
as functions of 
. Recall that a curve in the space can be written in parametric form in terms of only one variable. In this case we are considering the ‘‘natural’’ parametrization 
.
Recall that the parametric equation of a line has the form
,
where 
is a point on the line (in this particular case is (3,4,9)) and 
is the direction vector of the line. In this case, the direction vector of the line is the tangent vector of the ellipse at the point (3,4,9).
Now, if we have the parametric equation of a curve its tangent line will have direction vector 
. So, as we need to calculate the equation of the tangent line at the point 
, we must obtain the tangent vector 
. This part can be done taking implicit derivatives in the systems that defines the ellipse.
So, let us write the system as
.
Then, taking implicit derivatives:
.
Now we substitute the values 
and 
, and we get the system of linear equations
,
where the unknowns are 
and 
.
The system is
,
and its solutions are
and 
.
Then, the direction vector of the tangent is
.
Finally, the tangent line has parametric equation
where 
.
