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Find the points on the hyperboloid $9x^2- 45y^2 + 5z^2 - 45$ where the tangent plane is parallel to the plane $x+5y-2z = 7$?

Can anyone help me figure this one out?
So far, I've figured out the gradient of the hyperboloid but I'm not sure where to go from there..

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Is the equation of the hyperboloid $9x^2- 45y^2 + 5z^2 - 45=0$? –  Américo Tavares Oct 16 '11 at 22:00

2 Answers 2

Step 1: Find the equation of the vector plane giving the direction of the tangent plane at the point $(a,b,c)$ of the hyperboloid. This should read as $\alpha x+\beta y+\gamma z=0$ for some $\alpha$, $\beta$ and $\gamma$ depending on $a$, $b$ and $c$. Step 2: Note that the vector plane $\alpha x+\beta y+\gamma z=0$ and the affine plane $x+5y-2z=7$ are parallel if and only if their normal vectors are colinear and compute some respective normal vectors. Step 3: Conclude.

(Check: $(a,b,c)=(\frac54,-\frac54,-\frac92)$.)

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Two planes are parallel if their normal vectors are parallel. That is, their gradients are scalar multiples of one another.

Let's denote the hyperboloid as $f(x,y,z) = 9x^2-45y^2+5z^2-45$ and the plane as $g(x,y,z) = x+5y-2z=7$.

Taking the gradient of each of these,

$$ \nabla f = \left< 18x, -90y, 10z \right> $$

$$ \nabla g = \left< 1, 5, -2 \right> $$

This yields the following equations (setting the components of each gradient equal to one another):

$$ 18x = 1 $$

$$ -90y = 5 $$

$$ 10z = -2 $$

Solving these gives the point $\left(\frac{1}{18}, -\frac{1}{18}, -\frac{1}{5}\right)$.

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