# Sketch the contour plot and the graph of the function: $f(x,y) = \sqrt{36 - 9x^2 - 4y^2}$

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.

• Do you mean "contour" plots? – 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.
• Do you agree that $9x^2=\frac{x^2}{\frac1{3^2}}$? – Siong Thye Goh Oct 31 '17 at 3:21
• 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. – Siong Thye Goh Oct 31 '17 at 3:58