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Find the semi-axes of the ellipse of intersection, formed when the plane $z = x\!+\!y$ intersects the ellipsoid $$\frac{x^2}{4}+\frac{y^2}{5}+\frac{z^2}{25} = 1.$$
Let $(x,\,y,\,z)$ be any point of the ellipsoid. The square $x^2\!+\!y^2\!+\!z^2$ of the distance of this point from the midpoint $(0,\,0,\,0)$ has under the constraints
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(1) |
the minimum and maximum values at the end points of the semi-axes of the ellipse. Since we have two constraints, we must take equally many Lagrange multipliers, $\lambda$ and $\mu$ . A necessary condition of the extremums of $$f \;:=\, x^2\!+\!y^2\!+\!z^2$$ is that in addition to (1), also the equations
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(2) |
are satisfied. I.e., we have five equations (1), (2) and five unknowns $\lambda$ , $\mu$ , $x$ , $y$ , $z$ .
The equations (2) give $$x \;=\; -\frac{2\mu}{\lambda\!+\!4}, \quad y \;=\; -\frac{5\mu}{2\lambda\!+\!10}, \quad z \;=\; \frac{25\mu}{2\lambda\!+\!50},$$ which expressions may be put into the equation $h = 0$ , and so on. One obtains the values $$\lambda_1 = -10, \quad \lambda_2 = -\frac{75}{17}, \quad \mu_1 = \pm6\sqrt{\frac{5}{19}}, \quad \mu_2 = \pm\frac{140}{17\sqrt{646}}$$ with which the extremal points $(x,\,y,\,z)$ can be evaluated. The corresponding values of $f$ are 10 and $\frac{75}{17}$ , whence the major semi-axis is $\sqrt{10}$ and the minor semi-axis $\frac{5\sqrt{255}}{17}$ .
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