rhedynogh0rp
2022-03-25
Answered

Convert hyperbola in rectangular form to polar form

$3{y}^{2}-16y-{x}^{2}+16=0.$

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Mason Knight

Answered 2022-03-26
Author has **11** answers

Step 1

Staring with

$3{y}^{2}-16y-{x}^{2}+16=0$

Put it first in the standard format. Compute the square in y

$3{(y-d\frac{8}{3})}^{2}-d\frac{64}{3}-{x}^{2}+16=0$

$3{(y-d\frac{8}{3})}^{2}-{x}^{2}=d\frac{16}{3}$

$d\frac{9}{16}{(y-d\frac{8}{3})}^{2}-d\frac{3}{16}{x}^{2}=1$

Thus the center is at

$(0,\text{}\frac{8}{3}),\text{}a=\frac{4}{3}$ 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 $(0,0)$ and $(0,\frac{16}{3})$

Taking the focus that is at the origin, then $x=r\mathrm{cos}\theta ,\text{}y=r\mathrm{sin}\theta$

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

${r}^{2}(4{\mathrm{sin}}^{2}\theta -1)-16r\mathrm{sin}\theta +16=0$

From the quadratic formula, and taking the positive root

$r=d\frac{1}{2(4{\mathrm{sin}}^{2}\theta -1)}(16\mathrm{sin}\theta -\sqrt{256{\mathrm{sin}}^{2}\theta -(256{\mathrm{sin}}^{2}\theta -64)})$

And this simplifies to

$r=d\frac{1}{2(4{\mathrm{sin}}^{2}\theta -1)}(16\mathrm{sin}\theta -8)$

Factoring the denominator and cancelling equal terms

$r=d\frac{4}{2\mathrm{sin}\theta +1}$

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