How to Parametrize an Ellipse in Cartesian Coordinates?

[mathmari]

Hey!

Find the curve $\overrightarrow{\sigma}(t)$ that describes the following curve or trajectory. Make a graph.

$$\{(x, y) \mid 4x^2+y^2=1\}$$

How can I find such a curve??

 

[I like Serena]

Hey!

Find the curve $\overrightarrow{\sigma}(t)$ that describes the following curve or trajectory. Make a graph.

$$\{(x, y) \mid 4x^2+y^2=1\}$$

How can I find such a curve??

Hi! :)

Let's start with taking a look at its graph.
What does it look like? (Wondering)

 

[mathmari]

Let's start with taking a look at its graph.
What does it look like? (Wondering)

It is an ellipse, or not?? (Wondering)

 

[I like Serena]

It is an ellipse, or not?? (Wondering)

Yep. It's an ellipse. (Nod)

Do you know a parametric form of an ellipse? (Wondering)

Hint: it's similar to the form of a circle.

 

[mathmari]

Yep. It's an ellipse. (Nod)

Do you know a parametric form of an ellipse? (Wondering)

Hint: it's similar to the form of a circle.

It is $$x=a \cos t \\ y=b \sin t$$ right?? (Wondering)

 

[I like Serena]

It is $$x=a \cos t \\ y=b \sin t$$ right?? (Wondering)

Yep.

That leaves figuring out what $a$ and $b$ are... (Thinking)

 

[mathmari]

That leaves figuring out what $a$ and $b$ are... (Thinking)

Since $$4x^2+y^2=1 \Rightarrow 4a^2\cos^2 t+b^2\sin^2=\cos^2 t+\sin^2 t\Rightarrow 4a^2=1 \text{ AND } b^2=1 \Rightarrow a=\pm \frac{1}{2} \text{ AND } b=\pm 1$$

So, $$x=\frac{1}{2}\cos t \\ y=\sin t$$

Is this correct?? (Wondering)

 

[soroban]

Hello, mathmari!

Find the curve $\overrightarrow{\sigma}(t)$ that describes the following curve or trajectory.
Make a graph.

$$\{(x, y) \mid 4x^2+y^2=1\}$$

How can I find such a curve?

Do you recognize the equation of an ellipse?

We have: $\displaystyle\:\frac{x^2}{\frac{1}{4}} + \frac{y^2}{1} \:=\:1 \quad\Rightarrow\quad \frac{x^2}{(\frac{1}{2})^2} + \frac{y^2}{1^2} \:=\:1$

The ellipse is centered at the Origin.
Its vertices are: $\: (\pm\frac{1}{2},0)\,\text{ and }(0,\pm1)$

 

[I like Serena]

Since $$4x^2+y^2=1 \Rightarrow 4a^2\cos^2 t+b^2\sin^2=\cos^2 t+\sin^2 t\Rightarrow 4a^2=1 \text{ AND } b^2=1 \Rightarrow a=\pm \frac{1}{2} \text{ AND } b=\pm 1$$

So, $$x=\frac{1}{2}\cos t \\ y=\sin t$$

Is this correct?? (Wondering)

Yep.

And as Soroban already observed, it shows yet again that the semi axes are $\frac 1 2$ respectively $1$. ;)