Math Problem Statement

Find a point on the hyperboloid x2 + 4y2 − z2 = 1 where the tangent plane is parallel to the plane x + 4y − z = 0.

Solution

To find a point on the hyperboloid x2+4y2z2=1x^2 + 4y^2 - z^2 = 1 where the tangent plane is parallel to the plane x+4yz=0x + 4y - z = 0, follow these steps:

Step 1: Gradient of the Hyperboloid

The equation of the hyperboloid is: f(x,y,z)=x2+4y2z21=0f(x, y, z) = x^2 + 4y^2 - z^2 - 1 = 0 The normal vector to the surface at any point is given by the gradient of the function: f(x,y,z)=(fx,fy,fz)\nabla f(x, y, z) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) Calculating the partial derivatives: fx=2x,fy=8y,fz=2z\frac{\partial f}{\partial x} = 2x, \quad \frac{\partial f}{\partial y} = 8y, \quad \frac{\partial f}{\partial z} = -2z Thus, the gradient is: f(x,y,z)=(2x,8y,2z)\nabla f(x, y, z) = (2x, 8y, -2z) This vector is normal to the tangent plane at any point on the hyperboloid.

Step 2: Normal Vector to the Given Plane

The equation of the given plane is x+4yz=0x + 4y - z = 0. The normal vector to this plane is simply the coefficients of xx, yy, and zz: n=(1,4,1)\mathbf{n} = (1, 4, -1)

Step 3: Tangent Plane Condition

For the tangent plane to the hyperboloid to be parallel to the given plane, their normal vectors must be proportional. Therefore, the gradient (2x,8y,2z)(2x, 8y, -2z) should be a scalar multiple of (1,4,1)(1, 4, -1). This gives the system of equations: 2x=λ1,8y=λ4,2z=λ(1)2x = \lambda \cdot 1, \quad 8y = \lambda \cdot 4, \quad -2z = \lambda \cdot (-1) Simplifying these equations: 2x=λ,8y=4λ,2z=λ2x = \lambda, \quad 8y = 4\lambda, \quad -2z = \lambda From the first and third equations, we get: x=λ2,z=λ2x = \frac{\lambda}{2}, \quad z = \frac{\lambda}{2} From the second equation: 8y=4λy=λ28y = 4\lambda \quad \Rightarrow \quad y = \frac{\lambda}{2}

Step 4: Substituting into the Hyperboloid Equation

Now substitute x=λ2,y=λ2,z=λ2x = \frac{\lambda}{2}, y = \frac{\lambda}{2}, z = \frac{\lambda}{2} into the hyperboloid equation x2+4y2z2=1x^2 + 4y^2 - z^2 = 1: (λ2)2+4(λ2)2(λ2)2=1\left( \frac{\lambda}{2} \right)^2 + 4\left( \frac{\lambda}{2} \right)^2 - \left( \frac{\lambda}{2} \right)^2 = 1 Simplifying: λ24+4λ24λ24=1\frac{\lambda^2}{4} + 4 \cdot \frac{\lambda^2}{4} - \frac{\lambda^2}{4} = 1 [ \frac{\lambda^2}{4} + \lambda^2 - \frac{\lambda

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Math Problem Analysis

Mathematical Concepts

Multivariable Calculus
Gradient
Tangent Planes
Hyperboloids

Formulas

Gradient of f(x, y, z): ∇f(x, y, z) = (∂f/∂x, ∂f/∂y, ∂f/∂z)
Tangent plane condition: ∇f(x, y, z) = λ * normal vector of the given plane

Theorems

Condition for Parallel Planes (normal vectors must be proportional)

Suitable Grade Level

Undergraduate Mathematics (Multivariable Calculus)