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


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

### Step 1: Gradient of the Hyperboloid
The equation of the hyperboloid is:
\[
f(x, y, z) = x^2 + 4y^2 - z^2 - 1 = 0
\]
The normal vector to the surface at any point is given by the gradient of the function:
\[
\nabla f(x, y, z) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right)
\]
Calculating the partial derivatives:
\[
\frac{\partial f}{\partial x} = 2x, \quad \frac{\partial f}{\partial y} = 8y, \quad \frac{\partial f}{\partial z} = -2z
\]
Thus, the gradient is:
\[
\nabla f(x, y, z) = (2x, 8y, -2z)
\]
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 \( x + 4y - z = 0 \). The normal vector to this plane is simply the coefficients of \(x\), \(y\), and \(z\):
\[
\mathbf{n} = (1, 4, -1)
\]

### 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 \( (2x, 8y, -2z) \) should be a scalar multiple of \( (1, 4, -1) \). This gives the system of equations:
\[
2x = \lambda \cdot 1, \quad 8y = \lambda \cdot 4, \quad -2z = \lambda \cdot (-1)
\]
Simplifying these equations:
\[
2x = \lambda, \quad 8y = 4\lambda, \quad -2z = \lambda
\]
From the first and third equations, we get:
\[
x = \frac{\lambda}{2}, \quad z = \frac{\lambda}{2}
\]
From the second equation:
\[
8y = 4\lambda \quad \Rightarrow \quad y = \frac{\lambda}{2}
\]

### Step 4: Substituting into the Hyperboloid Equation
Now substitute \( x = \frac{\lambda}{2}, y = \frac{\lambda}{2}, z = \frac{\lambda}{2} \) into the hyperboloid equation \( x^2 + 4y^2 - z^2 = 1 \):
\[
\left( \frac{\lambda}{2} \right)^2 + 4\left( \frac{\lambda}{2} \right)^2 - \left( \frac{\lambda}{2} \right)^2 = 1
\]
Simplifying:
\[
\frac{\lambda^2}{4} + 4 \cdot \frac{\lambda^2}{4} - \frac{\lambda^2}{4} = 1
\]
\[
\frac{\lambda^2}{4} + \lambda^2 - \frac{\lambda

Find a point on the hyperboloid x^2 + 4y^2 − z^2 = 1 where the tangent plane is parallel to the plane x + 4y − z = 0.

Learn how to find a point on the hyperboloid x^2 + 4y^2 − z^2 = 1 where the tangent plane is parallel to the plane x + 4y − z = 0. This problem explores multivariable calculus concepts, including gradients, tangent planes, and hyperboloids. Suitable for undergraduate mathematics students.

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