# Convert hyperbola in rectangular form to polar form

Convert hyperbola in rectangular form to polar form
$3{y}^{2}-16y-{x}^{2}+16=0.$
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Mason Knight

Step 1
Staring with
$3{y}^{2}-16y-{x}^{2}+16=0$
Put it first in the standard format. Compute the square in y
$3{\left(y-d\frac{8}{3}\right)}^{2}-d\frac{64}{3}-{x}^{2}+16=0$
$3{\left(y-d\frac{8}{3}\right)}^{2}-{x}^{2}=d\frac{16}{3}$
$d\frac{9}{16}{\left(y-d\frac{8}{3}\right)}^{2}-d\frac{3}{16}{x}^{2}=1$
Thus the center is at
and $b=\frac{4}{\sqrt{3}}$
The focal distance
$c=\sqrt{{a}^{2}+{b}^{2}}=4\sqrt{d\frac{1}{9}+d\frac{1}{3}}=d\frac{8}{3}$
Hence, the foci are at $\left(0,0\right)$ and $\left(0,\frac{16}{3}\right)$
Taking the focus that is at the origin, then
Back to the original equation
$3{y}^{2}-16y-{x}^{2}+16=0$
Substituting the polar expressions,
$3{\left(r\mathrm{sin}\theta \right)}^{2}-16r\mathrm{sin}\theta -{r}^{2}{\mathrm{cos}}^{2}\theta +16=0$
Using ${\mathrm{cos}}^{2}\theta =1-{\mathrm{sin}}^{2}\theta$ and collecting terms
${r}^{2}\left(4{\mathrm{sin}}^{2}\theta -1\right)-16r\mathrm{sin}\theta +16=0$
From the quadratic formula, and taking the positive root

And this simplifies to
$r=d\frac{1}{2\left(4{\mathrm{sin}}^{2}\theta -1\right)}\left(16\mathrm{sin}\theta -8\right)$
Factoring the denominator and cancelling equal terms
$r=d\frac{4}{2\mathrm{sin}\theta +1}$