Complete question is;
The paraboloid z = 5 − x − x² − 2y² intersects the plane x = 1 in a parabola. Find parametric equations for the tangent line to this parabola at the point (1, 2, -4).
Answer:
x = 1
y = (2 + t)
z = (-4 - 8t)
Step-by-step explanation:
We are given the paraboloid z = 5 − x − x² − 2y²
That it intersects the plane x = 1 and tangent to the point (1, 2, -4).
Thus, the position of the tangent is;
r_o = (1, 2, -4).
Direction vector is; v = (x, y, z)
Let's put 1 for x into the paraboloid.
z = 5 − 1 − 1² − 2y²
z = 3 - 2y²
Now, let's use vectors to find the parametric equation, where;
x = 1
y = t
z = 3 - 2t²
Thus, the Directional vector is now;
v = (1, t, 3 - 2t²)
Derivative of this directional vector is;. v' = (0, 1, -4t)
Putting 2 for t gives;
v' = (0, 1, -4(2))
v' = (0, 1, -8)
Multiplying this derivative by T gjves;
tv' = (0, t, -8t)
The final direction vector will be;
r = r_o + tv'
r = (1, 2, -4) + (0, t, -8t)
r = (1, (2 + t), (-4 - 8t))
Thus our parametric equation is;
x = 1
y = (2 + t)
z = (-4 - 8t)
