Question 1033703
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Convert the polar equation to cartesian/rectangular coordinates

(r^2) cos (2 theta) = 2 

Thank you!
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<pre>
Let's do it together.
We start with the point ({{{r*cos(theta)}}},{{{r*sin(theta)}}}) in a coordinate plane. This point is presented in the polar form, 
and we want to learn what is a curve  {{{(r^2)*cos(2*theta)}}} = {{{2}}}.

Do you know that {{{cos(2theta)}}} = {{{cos^2(theta) - sin^2(theta)}}} ?

It is the direct consequence of the addition formula for cosine.
Also, it is named the formula for double argument for cosine.
If you don't know it, then look in these lessons 

   <A HREF=https://www.algebra.com/algebra/homework/Trigonometry-basics/Addition-and-subtraction-formulas.lesson>Addition and subtraction formulas</A>
   <A HREF=https://www.algebra.com/algebra/homework/Trigonometry-basics/Trigonometric-functions-of-multiply-argument.lesson>Trigonometric functions of multiply argument</A>

OK. Having this, let's make one step further.

{{{r^2*cos(2theta)}}} = {{{r^2*(cos^2(theta) - sin^2(theta))}}} = {{{r*(cos(theta)+sin(theta))}}}.{{{r*(cos(theta)-sin(theta))}}} = {{{(r*cos(theta) + r*sin(theta))}}}.{{{(r*cos(theta) - r*sin(theta))}}}.   (1)

Now,  {{{r*cos(theta)}}}  is x-component of the point  ({{{r*cos(theta)}}}, {{{r*sin(theta)}}})  in the rectangular coordinate system on the coordinate plane. 
So we can write {{{x}}} = {{{r*cos(theta)}}}.

Similarly,  {{{r*sin(theta)}}}  is y-component of the point  ({{{r*cos(theta)}}},{{{r*sin(theta)}}})  in the rectangular coordinate system on the coordinate plane. 
So we can write {{{y}}} = {{{r*sin(theta)}}}.

Therefore, we can rewrite the formula (1) in this way:

{{{r^2*cos(2theta)}}} = {{{(r*cos(theta) + r*sin(theta))}}}.{{{(r*cos(theta) - r*sin(theta))}}} = {{{(x+y)*(x-y)}}} =  {{{x^2 - y^2}}}.     (3)

Then your original equation  {{{r^2(cos(2*theta))}}} = {{{2}}}  becomes 
{{{x^2 - y^2}}} = {{{2}}}.

Do you recognize this equation ?  What is this ?  Is it familiar to you ?


But of course, it is the equation of a hyperbola . . . 


So, we unraveled this mystery.
</pre>

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Comment from student: It looks like the equation for a circle to me. A circle with a diameter of 2 centered at (0,0), correct? Thank you for your help!
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<U>My response</U>:  Do not make a mistake !


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;It is definitely the equation of a HYPERBOLA.


About equations for hyperbola see the lesson <A HREF=https://www.algebra.com/algebra/homework/Quadratic-relations-and-conic-sections/Hyperbola-definition--canonical-equation--characteristic-points-and-elements.lesson>Hyperbola definition, canonical equation, characteristic points and elements</A> in this site.