# Manipulating a polar equation

I started with this:

$$\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$$

And via substitution, got this far:

$$r=\left(\frac{144}{9-25\sin^{2}\theta}\right)^{.5}$$

For the fact that Desmos plots these the same, I assume I'm right so far.

The goal in this section is to end with the form

$$r=\frac{ep}{1-e\sin\theta}$$

and I am at a loss what the next manipulation might be.

• That type of form occurs when the center of the polar coordinates is one of the foci of the hyperbola. Commented Nov 15, 2021 at 20:01
• The pole of the polar equation in the desired equation is not the center of hyperbola, but one of the foci. Commented Nov 15, 2021 at 20:01
• @Potato Could you possibly explain how this helps? I don't study polar equations Commented Nov 15, 2021 at 20:16
• @random - Are you suggesting I first shift the equation to set the focus to (0,0)? If that's the case, this is perfect. Will look to do now as I await a reply. Commented Nov 15, 2021 at 20:21
• Yes. Exactly. Shift one of the foci (say (5, 0) ) to (0,0), then write the equation in polar coordinates. Commented Nov 15, 2021 at 20:59

The right focus is $$(5, 0)$$. Shifting the hyperbola to the left by $$5$$ units results in the new equation:

$$\dfrac{(x + 5)^2}{16} - \dfrac{y^2}{9} = 1$$

Now using polar coordinates, but measuring the angle $$\theta$$ from the positive $$y$$ axis direction, then $$x = -r \sin \theta , y = r \cos \theta$$

Hence,

$$\dfrac{ (-r \sin \theta + 5 )^2 }{16} - \dfrac{ (r \cos \theta)^2 }{9} = 1$$

Multiplying through by $$144$$,

$$9 ( r^2 \sin^2 \theta - 10 r \sin \theta + 25 ) - 16 r^2 \cos^2 \theta = 144$$

Simplifiying,

$$r^2 ( 25 \sin^2 \theta - 16 ) - 90 r \sin \theta + 81 = 0$$

$$r = \dfrac{1}{ 2(25 \sin^2 \theta - 16) } \left( 90 \sin \theta - \sqrt{ 8100 \sin^2 \theta - (8100 \sin^2 \theta - 5184 ) } \right)$$

And this simplifies to,

$$r = \dfrac{ 90 \sin \theta -72 }{2(5 \sin \theta - 4)(5 \sin \theta + 4) }$$

which simplifies further to

$$r = 9 \dfrac{ 5 \sin \theta - 4 }{ (5 \sin \theta - 4)(5 \sin \theta + 4 ) } = \dfrac{ 9 }{ 5 \sin \theta + 4 }$$

$$e = \sqrt{1 + \dfrac{3}{4}^2 } = \dfrac{5}{4}$$

Therefore,

$$r = \dfrac{ 9/4 }{ 1 + e \sin \theta }$$

• Your method spun the orientation 90 degrees. But, following your suggestion, relocating the focus at origin, was the answer I needed. I got $r=\frac{\frac{9}{4}}{1+\frac{5}{4}\cos\theta}$ which matched the graph perfectly. Much thanks. Commented Nov 15, 2021 at 23:52