Question

Are there any points on the hyperboloid \({x^2} - {y^2} - {z^2} = 1\) where the tangent plane is parallel to the plane \(z = x + y\) ?

Short answer

There is no point on the hyperboloid.

Step-by-step solution

Condition for two planes to be parallel.

For two planes to be parallel, their normal vectors must be parallel. The normal vector to the tangent plane of the hyperboloid at the point\((x,y,z)\)is

\({{\bf{n}}_1} = \nabla F(x,y,z) = \langle 2x, - 2y, - 2z\rangle \)

Find if there any points on the hyperboloid \({x^2} - {y^2} - {z^2} = 1\) where the tangent plane is parallel to the plane \(z = x + y\) 

It is given that,

The equation of the hyperboloid is, \({x^2} - {y^2} - {z^2} = 1\).

The equation of the plane is, \(z = x + y\).

The normal vector to the hyperboloid is, \(F(x,y,z) = {x^2} - {y^2} - {z^2} - 1\).

\(\begin{aligned}{l}\nabla F(x,y,z) = \left\langle {{F_x},{F_y},{F_z}} \right\rangle \\ = \left\langle {\frac{\partial }{{\partial x}}\left( {{x^2} - {y^2} - {z^2} - 1} \right),\frac{\partial }{{\partial y}}\left( {{x^2} - {y^2} - {z^2} - 1} \right),\frac{\partial }{{\partial z}}\left( {{x^2} - {y^2} - {z^2} - 1} \right)} \right\rangle \\ = \langle (2x),( - 2y),( - 2z)\rangle \end{aligned}\)

Thus, normal vector of the hyperboloid is \(\nabla F(x,y,z) = \langle 2x, - 2y, - 2z\rangle \).

The plane equation can be rewritten as, \(x + y - z = 0\).

The tangent plane is parallel to the plane equation \(x + y - z = 0\) only if the normal vectors of the planes are parallel. Thus, find the point on the hyperboloid such that \(\nabla F(x,y,z) = k\langle 1,1, - 1\rangle \). Then,

\(\begin{aligned}{l}\nabla F(x,y,z) = k\langle 1,1, - 1\rangle \\\langle 2x, - 2y, - 2z\rangle = k\langle 1,1, - 1\rangle \end{aligned}\)

Compare the above equations,

\(\begin{aligned}{l}2x = k, - 2y = k, - 2z = - k\\x = \frac{k}{2},y = - \frac{k}{2},z = \frac{k}{2}\end{aligned}\)

Substitute the values of x, y, z in \({x^2} - {y^2} - {z^2} = 1\),

\(\begin{aligned}{l}{\left( {\frac{k}{2}} \right)^2} - {\left( { - \frac{k}{2}} \right)^2} - {\left( {\frac{k}{2}} \right)^2} = 1\\\frac{{{k^2}}}{4} - \frac{{{k^2}}}{4} - \frac{{{k^2}}}{4} = 1\\\frac{{{k^2} - {k^2} - {k^2}}}{4} = 1\\\frac{{{k^2} - 2{k^2}}}{4} = 1\end{aligned}\)

Simplify further as follows,

\(\begin{aligned}{l}\frac{{ - {k^2}}}{4} = 1\\ - {k^2} = 4\\{k^2} = - 4\end{aligned}\)

There is no real solution and there is no point on the hyperbola.

Therefore, there are no points on the tangent plane.