Find the equation of the tangent plane and symmetric equation for the normal line to the given surface at P:
xy^2 + zy^2 + 4y -xz^2 = 18
P(-2,0,3)
To find the equation of the tangent plane to a surface at a given point P, we need to find the normal vector to the surface at that point. Then, we can use the general equation of a plane to write the equation of the tangent plane.
Step 1: Find the gradient vector of the given surface.
The gradient vector at any point represents the direction of the normal to the surface at that point. To find the gradient vector, we need to compute the partial derivatives of the surface equation with respect to each variable (x, y, and z).
Given surface equation: xy^2 + zy^2 + 4y - xz^2 = 18
∂/∂x (xy^2 + zy^2 + 4y - xz^2) = y^2 - z^2
∂/∂y (xy^2 + zy^2 + 4y - xz^2) = 2xy + 2zy + 4
∂/∂z (xy^2 + zy^2 + 4y - xz^2) = 2zy - 2xz
So, the gradient vector is given by:
∇f(x,y,z) = (y^2 - z^2, 2xy + 2zy + 4, 2zy - 2xz)
Step 2: Evaluate the gradient vector at the given point P(-2, 0, 3).
Substitute the x, y, and z values of P into the gradient vector obtained in step 1.
∇f(-2, 0, 3) = ((0)^2 - (3)^2, 2(-2)(0) + 2(0)(3) + 4, 2(3)(0) - 2(-2)(-2))
Simplifying, we get ∇f(-2, 0, 3) = (-9, 4, 4)
Step 3: Write the equation of the tangent plane using the normal vector.
Since the normal vector represents the direction of the normal to the plane, we can write the equation of the tangent plane in the form:
A(x - x_0) + B(y - y_0) + C(z - z_0) = 0
Using the values of the normal vector and the given point P(-2, 0, 3), we can write the equation of the tangent plane as:
-9(x - (-2)) + 4(y - 0) + 4(z - 3) = 0
Simplifying, we get -9x + 18 + 4y + 4z - 12 = 0
which can be further simplified as -9x + 4y + 4z + 6 = 0.
Thus, the equation of the tangent plane to the given surface at point P(-2, 0, 3) is -9x + 4y + 4z + 6 = 0.
Now, let's find the symmetric equation for the normal line.
Step 4: Write the parameterized equation for the normal line.
The parameterized equation for a line passing through a given point P and with direction vector parallel to the normal vector is given by:
x = x_0 + at
y = y_0 + bt
z = z_0 + ct
where (a, b, c) is the direction vector of the line.
We already have the given point P(-2, 0, 3) and the direction vector (-9, 4, 4). Substituting these values, we get the parameterized equation for the normal line as:
x = -2 - 9t
y = 0 + 4t
z = 3 + 4t
This is the parameterized equation for the normal line to the given surface at point P(-2, 0, 3).
Finally, the symmetric equation for the normal line can be obtained by eliminating the parameter t from the parameterized equations. In this case, we can rewrite the equations in terms of x, y, and z:
(x + 2) / -9 = (y - 0) / 4 = (z - 3) / 4
This is the symmetric equation for the normal line to the given surface at point P(-2, 0, 3).