So, yesterday…yes, Christmas…I proved (or perhaps, better said “derived”) the Quadratic Formula. The fun didn’t stop there. 🙂

My oldest son (visiting for the holiday), seemed to enjoy what I was doing…so I figured I’d blow his mind with how polar equations act in graphs, first showing him a simple \(r = 2\sin \theta\) one. After a bit discussing that, I thought I’d blow his mind more and create the equivalent of \(x = 1\), which is \(r = \frac{1}{{\cos \theta }}\).

Having a brain cramp, I thought I’d shift it one by adding 1 to it (\(r = \frac{1}{{\cos \theta }} + 1\)), but that results in something quite weird. You see, in a \(g(x) = f(x) + 1\) world, \(g(x)\) is just \(f(x)\) shifted up 1. Not so in the polar equation world. 🙂

The green line is \(r = \frac{1}{{\cos \theta }}\). The red lines are what happened when I added one to it (\(r = \frac{1}{{\cos \theta }} + 1\)):

But, why would we want to stop having fun there? Could we convert the polar equation to a Cartesian one? What would…

\[r = \frac{1}{{\cos \theta }} + 1\]

…be with \(x\) and \(y\)?

To help our transformation, let’s remember three things:

\[\begin{gathered}

{r^2} = {x^2} + {y^2} \hfill \\

x = r\cos \theta \hfill \\

y = r\sin \theta \hfill \\

\end{gathered} \]

Ready?!

Start with our original:

\[r = \frac{1}{{\cos \theta }} + 1\]

Get rid of the \(cos()\) in the denominator by multiplying both sides by \(\cos \theta\):

\[(r = \frac{1}{{\cos \theta }} + 1) \times \cos \theta \]

Which, multiplied out, is:

\[r\cos \theta = 1 + \cos \theta \]

Remember this?:

\[x = r\cos \theta \]

That means we can immediately transform what we have to:

\[x = 1 + \cos \theta \]

That’s great! However, how do we get rid of that pesky \(\cos \theta\) on the right?

Well, you may recall that:

\[\cos = \frac{{adjacent}}{{hypotenuse}}\]

What value is *adjacent* in an xy world? \(x\)

What value is *hypotenuse*? \(r\)

But \(r\) doesn’t do us much good…since it is a polar value. However, do you remember this?:

\[{r^2} = {x^2} + {y^2}\]

Therefore:

\[r = \sqrt {{x^2} + {y^2}} \]

(\(r\) is always positive.)

So…and it’s really about to get messy…replacing \(\cos \theta\) with *adjacent* divided by *hypotenuse* means:

\[x = 1 + \cos \theta \]

Becomes:

\[x = 1 + \frac{x}{{\sqrt {{x^2} + {y^2}} }}\]

Which is the same as:

\[x – 1 = \frac{x}{{\sqrt {{x^2} + {y^2}} }}\]

We want to end up with \(y\) alone on the left. To do that, we divide both sides by \(x-1\) and multiply both sides by \({\sqrt {{x^2} + {y^2}} }\), giving us:

\[\sqrt {{x^2} + {y^2}} = \frac{x}{{x – 1}}\]

Now we square both sides to get rid of the square root on the left:

\[{x^2} + {y^2} = {(\frac{x}{{x – 1}})^2}\]

Then, we move \(x^2\) to the right side of the equation by subtracting it from both sides:

\[{y^2} = {(\frac{x}{{x – 1}})^2} – x^2\]

And take the square root of each side so that it is \(y\) versus \(y^2\):

\[y = \pm \sqrt {{{(\frac{x}{{x – 1}})}^2} – {x^{^2}}} \]

Voila!

That was easy, eh? 🙂

*But, is it right?*

Let’s check it’s graph and see how it compares. The green line is \(x=1\) and the red is our new Cartesian equation:

Okay, to save you scrolling, here is our original beside it:

Did we do it?!

Well, it’s time to watch the rest of Die Hard 2. That’s a Christmas movie too, right? 🙂

P.S. You can see the polar graph here on Desmos, and the Cartesian one here.

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