Question

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To find the equation of the plane tangent to the surface

x^{2} + y^{2} - z^{2} + 29 = 0

at the given points

(6, 4, 9)

and

(- 4, - 6, 9)

, we can use the gradient vector.
The gradient vector of the surface at any point

(x, y, z)

is given by:

\nabla f (x, y, z) = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z})

In this case,

f (x, y, z) = x^{2} + y^{2} - z^{2} + 29

, so:

\nabla f (x, y, z) = (2 x, 2 y, - 2 z)

At the point

(6, 4, 9)

, the gradient vector is:

\nabla f (6, 4, 9) = (12, 8, - 18)

This vector is normal to the surface at the point

(6, 4, 9)

, so the equation of the plane tangent to the surface at that point is:

12 (x - 6) + 8 (y - 4) - 18 (z - 9) = 0

Simplifying this equation, we get:

6 x + 4 y - 9 z - 11 = 0

Similarly, at the point

(- 4, - 6, 9)

, the gradient vector is:

\nabla f (- 4, - 6, 9) = (- 8, - 12, - 18)

This vector is normal to the surface at the point

(- 4, - 6, 9)

, so the equation of the plane tangent to the surface at that point is:

- 8 (x + 4) - 12 (y + 6) - 18 (z - 9) = 0

Simplifying this equation, we get:

4 x + 6 y + 9 z + 117 = 0

Therefore, the equations of the planes tangent to the surface

x^{2} + y^{2} - z^{2} + 29 = 0

at the points

(6, 4, 9)

and

(- 4, - 6, 9)

are

6 x + 4 y - 9 z - 11 = 0

and

4 x + 6 y + 9 z + 117 = 0

, respectively.

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