MHB Convert polar equation sec(theta)=2 to rectangular equation

[Elissa89]

My professor gave us a study guide with the solutions:

The equation is:

sec(theta)=2

I am supposed to convert it to a rectangular equation. I know the answer is going to be y^2-3(x)^2=0

I don't know how to get to the answer he gave us.

 

[skeeter]

My professor gave us a study guide with the solutions:

The equation is:

sec(theta)=2

I am supposed to convert it to a rectangular equation. I know the answer is going to be y^2-3(x)^2=0

I don't know how to get to the answer he gave us.

note that $x = r\cos{\theta}$ and $r=\sqrt{x^2+y^2}$ ...$\sec{\theta} = \dfrac{1}{\cos{\theta}} = 2 \implies \cos{\theta}=\dfrac{1}{2}$

multiply both sides by $r$ ...

$r\cos{\theta} = \dfrac{r}{2} \implies x = \dfrac{\sqrt{x^2+y^2}}{2} \implies x^2 = \dfrac{x^2+y^2}{4}$$4x^2 = x^2+y^2 \implies y^2 - 3x^2 = 0$

 

[Elissa89]

note that $x = r\cos{\theta}$ and $r=\sqrt{x^2+y^2}$ ...$\sec{\theta} = \dfrac{1}{\cos{\theta}} = 2 \implies \cos{\theta}=\dfrac{1}{2}$

multiply both sides by $r$ ...

$r\cos{\theta} = \dfrac{r}{2} \implies x = \dfrac{\sqrt{x^2+y^2}}{2} \implies x^2 = \dfrac{x^2+y^2}{4}$$4x^2 = x^2+y^2 \implies y^2 - 3x^2 = 0$

*Sigh* It's always so obvious when someone shows me the steps but I have the hardest time figuring it out on my own. thanks