Question

The paraboloid z = 5 − x − x2 − 2y2 intersects the plane x = 1 in a parabola. Find parametric equations in terms of t for the tangent line to this par

Answer 1

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)