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.

  • 1
    $\begingroup$ Do you mean "contour" plots? $\endgroup$ – alex.jordan Oct 31 '17 at 4:12

$$36-k^2 = 9x^2+4y^2$$

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

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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.