I am having a bit of trouble beginning this question.
The question is as follows:
Let $r$ be the curve which is the intersection of the surface $z = \frac{1}{3}x^2 + \frac{2}{3}y^2$ and the surface: $x^2 + y^2 + z^2 = 3$.
Find the equation of the tangent line to the curve $r$ at $\overrightarrow{r_0} = (1, 1, 1)$, and the equation of the plane which is perpendicular to the tangent line at $\overrightarrow{r_0}$.
I think I understand how to approach the second part of the problem, but I'm not too sure how to parameterize the intersection $r$.
I've tried substituting $z$ from the first surface in to the second surface:
$$x^2 + y^2 + \left(\frac{1}{3}x^2 + \frac{2}{3}y^2\right)^2 = 3\\
x^2 + y^2 + \frac{1}{9}x^4 + \frac{4}{9}x^2 y^2 + \frac{4}{9}y^4 = 3$$
Let $X = x^2$ and $Y = y^2$,
$$X + Y + \frac{1}{9}X^2 + \frac{4}{9}XY + \frac{4}{9}Y^2 = 3$$
I tried to eliminate the $XY$ but then the whole thing just became a mess with imaginary numbers. How should I proceed? Any help would be greatly appreciated.
Best Answer
Hint:
I would not try to parameterize the curve, which is the intersection of an elliptic paraboloid and a sphere. Spherical or cylindrical coordinates will not work so well.
I would rather consider the normal vectors to both surfaces (gradient) and use the fact that the tangent vector is orthogonal to both.