Level Curves of T(x, y) and V(x, y) - Revisiting Ellipses

[cscott]

Homework Statement

I need to sketch level curves of $T(x, y) = 50(1 + x^2 + 3y^2)^{-1}$ and $V(x, y) = \sqrt{1 - 9x^2 -4y^2}$

The Attempt at a Solution

Is it correct that they are ellipses?

ie [tex]1 = \frac{9}{1 - c^2} x^2 + \frac{4}{1 - c^2}y^2[/itex] <br /> <br /> for V(x, y) = c = constant<br /> I feel so rusty <a href="https://www.physicsforums.com/insights/want-go-back-school/" class="link link--internal">going back to school</a> :s[/tex]

 

[HallsofIvy]

That's for V? Setting $V(x,y)= \sqrt{1- 9x^2- 4y^2}= c$, then $1- 9x^2- 4y^2= c^2$, $9x^2+ 4y^2= 1-c^2$,
$$\frac{9}{1- c^2}x^2+ \frac{4}{1-c^2}y^2= 1$$
just as you say. Yes, that's an ellipse. It might be easier to recognise if you wrote it
$$\frac{x^2}{\left(\frac{\sqrt{1-c^2}}{3}\right)^2}+ \frac{y^2}{\left(\frac{\sqrt{1-c^2}}{2}\right)^2}= 1$$
an ellipse with center at (0,0) and semi-axes of length
$$\frac{\sqrt{1-c^2}}{3}$$
and
$$\frac{\sqrt{1-c^2}}{2}$$

Similarly, $T(x,y)= 50(1+ x^2+ 3y^2)^{-1}= c$ gives $c(1+ x^2+ 3y^2)= 50$ so $1+ x^2+ 3y^2= 50/c$, $x^2+ 3y^2= (50/c- 1)$. Now divide both sides by 50/c- 1:
$$\frac{x^2}{50/c-1}+ \frac{y^2}{\frac{50/c-1}{3}}= 1$$
again, an ellipse with center at (0,0), semi-axes of length
$$\sqrt{50/c- 1}$$
and
$$\sqrt{\frac{50/c- 1}{3}}$$

 

[cscott]

That does make it easier to understand.

Thanks for your help.