0
$\begingroup$

I'm trying to find the points on the curve $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1 $ such that the normal to the curve forms equal angles with the coordinate axes. I treated the curve as a level set of the function of 3 variables $F(x,y,z)$ and so the curve is perpendicular to the gradient $\nabla F(x,y,z)=(\frac{2x}{a^2},\frac{2y}{b^2}, \frac{2z}{c^2}) $ but I don't really know where to go from here.

2 Answers 2

0

The gradient will have equal angles with respect to the coordinate axes when each angle is 45 degrees. This occurs when all components are equal so the set of points where all your components are equal is what you want. Setting each coordinate pairwise equal to the others gives you a system of three equations and three unknowns.

edit: If you think about just two dimensions, the x-component of a vector is equal to the y-component if the angle with respect to axes is 45 degrees. It's the same for 3 dimensions. In this case your three equations would be $$\frac{2x}{a^2}=\frac{2y}{b^2} \quad \frac{2y}{b^2} = \frac{2z}{c^2} \quad \frac{2x}{a^2} = \frac{2z}{c^2} $$

  • 0
    How do you set up the system of equations exactly? I'm not sure where the 3 equations come from. – 2017-02-13
  • 0
    Ok I made some edits that will hopefully help. – 2017-02-13
  • 0
    Would I then substitute these into the original equation of the curve? – 2017-02-13
  • 0
    Nope. Once you solve them like @Doug M did you are done. – 2017-02-13
0

If the angles between $\nabla F$ forms equal angles with the coordinate axes, then $\nabla F \cdot \mathbf i = \nabla F \cdot \mathbf j = \nabla F \cdot \mathbf k$

$\frac {x}{a^2} = \frac {y}{b^2} = \frac {z}{c^2}$

$\frac {x}{a} = \frac {a}{b} \frac {y}{b}\\ \frac {x^2}{a^2} = \frac {a^2}{b^2} \frac {y^2}{b^2}$

$(\frac {a^2}{b^2}+1) \frac {y^2}{b^2} + \frac {z^2}{c^2} = 1$

$\frac {y^2}{b^2} = \frac {b^2}{c^2} \frac {z^2}{c^2}$

$(\frac {a^2+b^2}{c^2} + 1) \frac {z^2}{c^2}= 1$

$z^2 = \frac {c^2}{a^2+b^2+c^2}\\ z = \pm \frac {c^2} {\sqrt {a^2+b^2+c^2}}$

similarly:

$x = \pm \frac {a^2} {\sqrt {a^2+ b^2+c^2}}\\ y = \pm \frac {b^2} {\sqrt {a^2 +b^2+ c^2}}$