At what point on the paraboloid

y = x2 + z2
is the tangent plane parallel to the plane
7x + 2y + 3z = 2?
(If an answer does not exist, enter DNE.)

Well, let's find the tangent plane to the paraboloid at a general point (x, y, z). The equation of the tangent plane is given by:

F(x, y, z) = f(a, b, c) + f_x(a, b, c)(x - a) + f_y(a, b, c)(y - b) + f_z(a, b, c)(z - c) = 0,

where f(x, y, z) = y - x^2 - z^2 is the equation of the paraboloid.

Taking partial derivatives of f, we get:

f_x(x, y, z) = -2x,
f_y(x, y, z) = 1,
f_z(x, y, z) = -2z.

Now we need a point (a, b, c) on the paraboloid that makes the tangent plane parallel to the given plane. The normal vectors of these two planes should be parallel, which means their direction ratios should be proportional.

The normal vector to the given plane 7x + 2y + 3z = 2 is <7, 2, 3>.

So we need to find a point (a, b, c) on the paraboloid such that the direction ratios of the normal vector to the tangent plane, which is <f_x(a, b, c), f_y(a, b, c), f_z(a, b, c)>, are proportional to <7, 2, 3>.

Let's equate the direction ratios:

-2x/7 = 1/2 = -2z/3.

From the first and second equations, we get x = -7/4 and z = -9/4, respectively.

Now, plugging these values into the equation of the paraboloid, we find:

y = (-7/4)^2 + (-9/4)^2 = 49/16 + 81/16 = 130/16 = 65/8.

Therefore, the point on the paraboloid where the tangent plane is parallel to the plane 7x + 2y + 3z = 2 is (-7/4, 65/8, -9/4).

To find the point on the paraboloid where the tangent plane is parallel to the given plane, we need to find the gradient vector of the paraboloid and the normal vector of the plane.

The equation of the paraboloid is y = x^2 + z^2. Let's write it as a function f(x, y, z) = x^2 + y - z^2 = 0.

Now, the gradient vector of f(x, y, z) at any point (x, y, z) is given by ∇f = (∂f/∂x, ∂f/∂y, ∂f/∂z).

Taking partial derivatives:
∂f/∂x = 2x
∂f/∂y = 1
∂f/∂z = -2z

So the gradient vector is ∇f = (2x, 1, -2z).

The equation of the plane is 7x + 2y + 3z - 2 = 0. We can rewrite it as g(x, y, z) = 7x + 2y + 3z - 2 = 0.

The normal vector of the plane is given by the coefficients of x, y, and z in the equation. So the normal vector of the plane is ∇g = (7, 2, 3).

For the tangent plane to be parallel to the given plane, the gradient vector of the paraboloid (∇f) must be parallel to the normal vector of the plane (∇g). This means that their corresponding components must be proportional.

Setting up proportions:
2x/7 = 1/2 = -2z/3

Simplifying each proportion:
2x = 7/2
-2z = 3/2

Solving each equation for x and z:
x = 7/4
z = -3/4

Now, we can substitute these values of x and z back into the equation of the paraboloid to find the corresponding value of y:
y = (7/4)^2 + (-3/4)^2 = 49/16 + 9/16 = 58/16 = 29/8

Therefore, the point on the paraboloid where the tangent plane is parallel to the given plane is (7/4, 29/8, -3/4).

To find the point on the given paraboloid where the tangent plane is parallel to the given plane, we need to determine the normal vector of the tangent plane.

The equation of the given paraboloid is y = x^2 + z^2, which can be written as f(x, y, z) = y - x^2 - z^2 = 0. The partial derivatives of f(x, y, z) with respect to x, y, and z are:

∂f/∂x = -2x
∂f/∂y = 1
∂f/∂z = -2z

The normal vector of the tangent plane to the paraboloid at a point (x0, y0, z0) is obtained by evaluating these partial derivatives at that point. In other words, the normal vector (a, b, c) is given by:

(a, b, c) = (∂f/∂x, ∂f/∂y, ∂f/∂z) evaluated at (x0, y0, z0)

We want the tangent plane to be parallel to the plane 7x + 2y + 3z = 2, which means the normal vectors of both planes should be proportional.

Comparing the components, we have the following system of equations:

a = -2x0
b = 1
c = -2z0

Comparing the ratios of the respective components:

-2x0 / 7 = 1 / 2 = -2z0 / 3

From the second fraction, we have 1 / 2 = -2z0 / 3, which implies z0 = -3/4.

Substituting z0 = -3/4 into the first fraction, we have -2x0 / 7 = -3/8. Solving this equation for x0 gives us x0 = 21/16.

We found x0 = 21/16 and z0 = -3/4. To find y0, we substitute these values into the equation of the paraboloid:

y0 = (21/16)^2 + (-3/4)^2 = 441/256 + 9/16 = 441/256 + 144/256 = 585/256

So, the point on the paraboloid where the tangent plane is parallel to the plane 7x + 2y + 3z = 2 is (x0, y0, z0) = (21/16, 585/256, -3/4).

In other words,

divide the plane's normal vector so that the y components match (INCLUDING SIGN), then solve for the two remaining variables.

For the example above this would mean;

2xi-j+2zk=7i+2j+3k
becomes
2xi-j+2xk=(-7/2)i-j+(-3/2)k

Now solve for x and z like so:
2x=(-7/2) => x=(-7/2)/2 = -7/4

2z=(-3/2) => z=(-3/2)/2 = -3/4

Now for y, just use the original equation (y=x^2+z^2), since we now know x and z, simply plug them in to get y like so:

y= (-7/4)^2+(-3/4)^2 = 29/8

So your coordinates are:

(-7/4, 29/8, -3/4)

Happy trails.

normal vector to f(x,y,z)=0 is

so, for the paraboloid,
n = fxi + fyj + fzk
= 2xi - j + 2zk

for the plane,
n = 7i + 2j + 3k

Now manipulate the directions and magnitudes so the two normals are the same.