Find equations of the tangent plane and normal line to the surface x=3y^2+3z^2−274 at the point (-4, -9, -3).

Question

Find equations of the tangent plane and normal line to the surface x=3y^2+3z^2−274 at the point (-4, -9, -3).

Tangent Plane: (make the coefficient of x equal to 1).

? =0

Normal line: ⟨−4, ? , ? ⟩ +t⟨1, ? , ? ⟩.

Note: Question marks are asked.

Answer 1

We rewrite this as x - 3y^2 - 3z^2 + 274 = 0

Then, with implicit differentation, we get

dx - 6ydy - 6z dz = 0

Thus, at (-4, -9, -3), as (1, -6y, -6z) = (1, 54,18),

our tangent plane is x + 54y + 18z = -4 + 54 * -9 + 18 * -3, or x + 54y + 18z = -544, or

x + 54y + 18z + 544 = 0,

and

our normal line is

(-4, -9, -3) + t(1, -6y, -6z) =

(-4, -9, -3) + t(1, -6(-9), -6(-3)) =

(-4, -9, -3) + t(1, 54, 18)

Answer 2

View the equation x - 3y² - 3z² + 274 = 0

as a level surface of the function F (x, y, z )= x - 3y^2 - 3z^2

Then its gradient vector at P ( -4, -9, -3 ) can serve us as the normal vector to the surface

and as the directing vector of the normal line at P.

Then ∇ F = [∂F / ∂x ] i + [∂F / ∂y ] j + [∂F / ∂z ] k = i - 6 y j -6 z k

and evaluated at ( -4, -9, -3) gives us the normal vector of the tangent plane

n = <1, 54, 18>

Then the equation of the tangent plane

is 1 (x+4) +54(y+9) +18(z+3) = 0

or equivalently

x + 54 y +18 z = -544

The parametric equations of the normal lie being

x= -4 + t

y= - 9 +54 t

z = -3 + 18 t