The paraboloid z = 6 − x − x2 − 5y2 intersects the plane x = 2 in a parabola. Find parametric equations in terms of t for the tangent line to this parabola at the point (2, 2, −20).

x = 2

y = 2 +  t

z = -20 -20t

Step-by-step explanation:

First, we are going to find the equation for this parabola. We replace x = 2 in the equation of the paraboloid, thus:

if x = 2, then

Now, we calculate the tangent line to this parabola at the point (2,2,-20)

The parametrization of the parabola is:

x = 2

y = t  

 since

We calculate the derivative

we evaluate the derivative in t=2, since at the point (2,2,-20) y = 2 and y = t

Thus:

Then, the director vector for the tangent line is (0,1,-20)

and the parametric equation for this line is:

x = 2

y = 2 +  t

z = -20 -20t

its kind of blurry. can you put a clearer pic?

step-by-step explanation:

step-by-step explanation:

points a and b are equidistant from the desired line, so the line you find by using them will go through point r.

using points other than a and b will give the desired line only if they are equidistant from r. if they are randomly located, the perpendicular line you create will not necessarily go through point r, hence will not be a solution to the problem.