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 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

 

[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.