0

I'm doing the exercise where given the function $$4x^2 + 9 y^2 = 1$$ I must describe how the level curves of that function will be.

I attended some classes on Conics for some time and did a little research before creating this question, but even so I can't understand why the graph has the points it has.

enter image description here

The points on the x and y axis are:

$$x =-\sqrt{1/4}$$ $$x =+\sqrt{1/4}$$ $$y =-\sqrt{1/9}$$ $$y =+\sqrt{1/9}$$

How do I get there?

Thiago
  • 141

2 Answers2

4

When I was typing this question, I found the solution. I just substituted $x = 0$ and $y = 0$ to find each one of the intersections.

Specifically, when $x = 0$:

$$9y^2 = 1 \Rightarrow y = \sqrt{\frac{1}{9}} = ±\frac{1}{3}$$

and when $y = 0$, $$4x^2 = 1 \Rightarrow x = \sqrt{\frac{1}{4}} = ±\frac{1}{2}.$$

Toby Mak
  • 16,827
Thiago
  • 141
  • 2
    Another way to see it is that the smallest $4x^2$ can be is $0$ and if $4x^2$ is the smallest it can be, $9y^2$ is the largest it can be which then gives that $y = \pm \dfrac{1}{\sqrt{9}} = \pm \dfrac{1}{3}$. And vice versa for $y$ and then $x$. – Cameron Williams Aug 27 '20 at 18:56
3

$$4x^2 + 9y^2 = 1$$ $$(2x)^2 + (3y)^2 = 1 $$ $$\left(\dfrac{x}{1/2}\right)^2 + \left(\dfrac{y}{1/3}\right)^2 = 1.$$

This equation is used for an ellipse with the width $2 \cdot 1/2$ and the height $2 \cdot 1/3$.

MachineLearner
  • 3,991