The plane $x – y + z = 2$ intersects the cylinder $x^2 + y^2 = 4$ in an ellipse. Find the points on the ellipse closest to and farthest from the origin.
Answer Farthest $(-\sqrt{2}, \;\sqrt{2},\; 2 + 2 × \sqrt{2})$
Answer Shortest $(2, 0, 0)$ and $(0, -2, 0)4
Here is my attempt:
since $$x^2 + y^2 = 4
y = \sqrt{4 – x^2 }$$
since $$x – y + z = 2\\x – \sqrt{4 – x^2} + z = 2,\\z = 2 – x + \sqrt{ 4 – x^2}$$
subtitute $x$ with $t$
now we have a parametric curve of the interection between two surfaces
$$r(t) = (t, \sqrt{ 4 – t^2 }, 2 – t + \sqrt{ 4 – t^2 })$$
Distance from the origin
$$D = \sqrt{( t^2 + (4 – t^2)} + ( 2 – t + \sqrt{4 – t^2})^2 )$$
Shortest Distance between the intersection and the origin
$$\dfrac{dr}{dt} = 0 = \dfrac{dr}{dt} [ \sqrt{ t^2 + (4 – t^2)} + ( 2 – t + \sqrt{4 – t^2}\;)^2 ) ]$$ … too much
Requesting help for finding a more simple way to solve this problem…
Best Answer
This can be done with Lagrange multipliers. We want to find the extrema of $f(x,y,z) = x^2+y^2+z^2$ given that $g(x,y,z) = x-y+z-2 = 0$ and that $h(x,y,z) = x^2+y^2-4 = 0$. Then we have $$\nabla f = \langle 2x,2y,2z \rangle$$ $$\nabla g = \langle 1,-1,1\rangle$$ $$\nabla h = \langle 2x, 2y, 0\rangle$$ Extrema satisfy the condition that $\nabla f = \mu \nabla g+\lambda \nabla h$ for some $\lambda, \mu \in \mathbb{R}$. This is to say, $$2x = 2\lambda x+\mu$$ $$2y = 2\lambda y-\mu$$ $$2z = \mu$$ If $\lambda = 1$ then $\mu = 0$ and so $z=0$. Using the $g$ constraint, we have that $x=y+2$, and so using the $h$ constraint, we have that $y^2+(y+2)^2=4$, which implies that $y=0$ or $y=2$. Along with the $g$ constraint, this gives us two critical points: $$(2, 0, 0)$$ $$(0, -2, 0)$$
Now assume $\lambda \neq 1$, and so $$x = \frac{\mu}{2-2\lambda}$$ $$y = \frac{-\mu}{2-2\lambda} = -x$$ Since $x=-y$, we have that $x=\pm \sqrt2$, $y = \mp \sqrt2$. Using the $g$ constraint, our two critical points are $$(\sqrt 2, -\sqrt 2, 2-2\sqrt 2)$$ $$(-\sqrt 2, \sqrt 2, 2+2\sqrt 2)$$ And then it's east to determine which is the max and which is the min out of these four critical points.