Question

The plane\(x + y + 2z = 2\)intersects the parabolic\(z = \mathop x\nolimits^2 + \mathop y\nolimits^2 \)in an ellipse. Find the points on this ellipse that are nearest to and farthest from the origin.

Short answer

The points on this ellipse that is nearest to and farthest from the origin.\(x + z = 1,\;\;\;2{x^2} = z\)

Step-by-step solution

Step-1

The key here is that we needn't worry about the equation of the ellipse formed by the intersection; instead, we regard the two surfaces as constraints and seek to minimize the distance from the origin. More precisely, we'll minimize

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

The distance squared, since this is minimized precisely when the distance is, and this function is a lot easier to work with. You should remember this trick. The constraints then are

\(g(x,y,z) = x + y + 2z - 2 = 0,\;\;\;h(x,y,z) = {x^2} + {y^2} - z = 0\)

We must now solve

\(\nabla f = \lambda \nabla g + \mu \nabla h\)

So we calculate,

\(\nabla f = \langle 2x,2y,2z\rangle ,\;\;\;\nabla g = \langle 1,1,2\rangle ,\;\;\;\nabla h = \langle 2x,2y, - 1\rangle \)

Step-2

And the equation we must solve becomes three equations, in addition to the two constraints.

\(2x = \lambda + 2x\mu ,\;\;\;2y = \lambda + 2y\mu ,\;\;\;2z = 2\lambda - \mu \)

Solving five equations for five unknowns can get tricky, to say the least. One idea is to observe that the first two equations above are quite similar. In fact, we can write

\(2x - 2x\mu = \lambda = 2y - 2y\mu (x - y)(2 - 2\mu ) = 0\)

And so either\(\mu = 1\)or\({\rm{x = y}}\)First consider the case where\(\mu = 1\)Then from the first equation above,\(\lambda = 0\)Then\(z = - \frac{1}{2}\)But then

\({x^2} + {y^2} = - \frac{1}{2}\)

And there are clearly no solutions in this case. So now assume\(x = y\). At this point, we can dispense with solving for\(\lambda \)and\(\mu \)entirely. (Remember, nothing about the problem insists we find their values!)So, going back to the original constraints, we have

\(x + z = 1,\;\;\;2{x^2} = z\)