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I've done a bit: range of function: $[0,6]$. So we need level curves for $k = 0,1,\ldots,6$ $$f(x,y) = k$$ $$\sqrt{36 - 9x^2 - 4y^2} = k$$ $$36 - 9x^2 - 4y^2 = k^2$$ $$36-k^2 = 9x^2 + 4y^2$$

Not really sure how to go further. I tried to divide by $9$ and $4$ to simplify.. but it didn't work out.

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    $\begingroup$ Do you mean "contour" plots? $\endgroup$ – alex.jordan Oct 31 '17 at 4:12
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$$36-k^2 = 9x^2+4y^2$$

$$1=\frac{x^2}{\left(\frac{\sqrt{36-k^2}}{3}\right)^2}+\frac{y^2}{\left( \frac{\sqrt{36-k^2}}{2}\right)^2}$$

Notice that $1=\frac{x^2}{a^2}+\frac{y^2}{b^2}$ is an equation of an ellipse.

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  • $\begingroup$ How did you get the denominators? Divided by 36-k^2 but what about the rest? $\endgroup$ – Math guy Oct 31 '17 at 3:15
  • $\begingroup$ Do you agree that $9x^2=\frac{x^2}{\frac1{3^2}}$? $\endgroup$ – Siong Thye Goh Oct 31 '17 at 3:21
  • $\begingroup$ Oh okay I see it now. And the 3 and 2 is the x and y intercepts? $\endgroup$ – Math guy Oct 31 '17 at 3:55
  • $\begingroup$ Consider the equation $1=\frac{x^2}{a^2}+\frac{y^2}{b^2}$, to find the $x$ intercept, let $y=0$, hence $x=\pm a$, hence $a$ and $-a$ are the $x$-intercept. Similarly for $y$-intercept. $\endgroup$ – Siong Thye Goh Oct 31 '17 at 3:58

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