# The plane x+y+2z=18 intersects the paraboloid z=x^2+y^2 in an ellipse.Find the points on this ellipse that are nearest to and farthest from the origin

The plane $$\displaystyle{x}+{y}+{2}{z}={18}$$ intersects the paraboloid $$\displaystyle{z}={x}^{{2}}+{y}^{{2}}$$ in an ellipse. Find the points on this ellipse that are nearest to and farthest from the origin.

• Questions are typically answered in as fast as 30 minutes

### Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

SoosteethicU

It suffices to find the extreme values of the square of the distance to the origin $$\displaystyle{D}={x}^{{2}}+{y}^{{2}}+{z}^{{2}}$$, subject to the constraints $$\displaystyle{g}={x}+{y}+{2}{z}={18}{\quad\text{and}\quad}{h}={x}^{{2}}+{y}^{{2}}-{z}={0}$$. $$\mu$$

By Lagrange Multipliers,$$\triangledown D = \lambda \triangledown g + \mu \triangledown h.$$ $$\displaystyle\Rightarrow$$$$<2x, 2y, 2z> = \lambda<1, 1, 2> + \mu<2x, 2y, -1>.$$

Equating like entries, $$\displaystyle{2}{x}=\lambda+{2}\mu{x}$$
$$\displaystyle{2}{y}=\lambda+{2}\mu{y}$$
$$\displaystyle{2}{z}={2}\lambda-\mu$$

Since the last equation can be rewritten as $$\displaystyle\mu={2}\lambda-{2}{z}$$, we now have $$\displaystyle{2}{x}=\lambda+{2}{x}{\left({2}\lambda-{2}{z}\right)}=\Rightarrow{2}{x}=\lambda{\left({4}{x}+{1}\right)}-{4}{x}{z}$$
$$\displaystyle{2}{y}=\lambda+{2}{y}{\left({2}\lambda-{2}{z}\right)}=\Rightarrow{2}{y}=\lambda{\left({4}{y}+{1}\right)}-{4}{y}{z}$$
So,$$\displaystyle\frac{{{2}{x}+{4}{x}{z}}}{{{4}{x}+{1}}}=\lambda=\frac{{{2}{y}+{4}{y}{z}}}{{{4}{y}+{1}}}\Rightarrow{2}{x}\frac{{{1}+{2}{z}}}{{{4}{x}+{1}}}={2}{y}\frac{{{1}+{2}{z}}}{{{4}{y}+{1}}}.$$

*If $$\displaystyle{1}+{2}{z}={0}\le\Rightarrow{z}=-\frac{{1}}{{2}},{t}{h}{e}{n}\ {h}={x}^{{2}}+{y}^{{2}}=-\frac{{1}}{{2}}$$ has no solution.

So, we can ignore this case.

Now, we can freely divide both sides by $$\displaystyle{\left({1}+{2}{z}\right)}$$:

$$\displaystyle\frac{{{2}{x}}}{{{4}{x}+{1}}}=\frac{{{2}{y}}}{{{4}{y}+{1}}}$$
$$\displaystyle\Rightarrow{2}{x}{\left({4}{y}+{1}\right)}={2}{y}{\left({4}{x}+{1}\right)}$$
$$\displaystyle\Rightarrow{x}={y}.$$

Substitute this into the constraint equations:
$$\displaystyle{x}+{x}+{2}{z}={18}\Rightarrow{x}+{z}={9}$$
$$\displaystyle{x}^{2}+{x}^{2}-{z}={0}\Rightarrow{z}={2}{x}^{2}$$

So,$$\displaystyle{x}+{2}{x}^{2}={9}.$$
$$\displaystyle\Rightarrow{2}{x}^{2}+{x}-{9}={0}$$
$$\displaystyle\Rightarrow{x}=\frac{{-{1}+\sqrt{{{73}}}}}{{4}}{\quad\text{or}\quad}\frac{{-{1}-\sqrt{{{73}}}}}{{4}}$$

So, $$\displaystyle{\left({x},{y}\right)}={\left(\frac{{-{1}+\sqrt{{{73}}}}}{{4}},\frac{{-{1}+\sqrt{{{73}}}}}{{4}}\right)}{\quad\text{or}\quad}{\left(\frac{{-{1}-\sqrt{{{73}}}}}{{4}},\frac{{-{1}-\sqrt{{{73}}}}}{{4}}\right)}$$,

since $$y = x$$
$$\displaystyle\Rightarrow{\left({x},{y},{z}\right)}={\left(\frac{{-{1}+\sqrt{{{73}}}}}{{4}},\frac{{-{1}+\sqrt{{{73}}}}}{{4}},\frac{{{37}-\sqrt{{{73}}}}}{{4}}\right)}{\quad\text{or}\quad}{\left(\frac{{-{1}-\sqrt{{{73}}}}}{{4}},\frac{{-{1}-\sqrt{{{73}}}}}{{4}},\frac{{{37}+\sqrt{{{73}}}}}{{4}}\right)}$$, using either constraint.
Finally, substitute this into D:
$$\displaystyle{D}{\left(\frac{{-{1}+\sqrt{{{73}}}}}{{4}},\frac{{-{1}+\sqrt{{{73}}}}}{{4}},\frac{{{37}-\sqrt{{{73}}}}}{{4}}\right)}=\frac{{{1590}-{78}\sqrt{{{3}}}}}{{16}}$$ <---Closest
$$\displaystyle{D}{\left(\frac{{-{1}-\sqrt{{{73}}}}}{{4}},\frac{{-{1}-\sqrt{{{73}}}}}{{4}},\frac{{{37}+\sqrt{{{73}}}}}{{4}}\right)}=\frac{{{1590}+{78}\sqrt{{{3}}}}}{{16}}$$ <---Farthest.