Nickolas Taylor

2022-07-12

While doing a sum I was stuck in a particular step:
${r}_{1}=\frac{4a\mathrm{cos}\theta }{{\mathrm{sin}}^{2}\theta }$
and
${r}_{2}=\frac{4a\mathrm{sin}\theta }{{\mathrm{cos}}^{2}\theta }$
How to eliminate $\theta$?

Expert

${r}_{1}\cdot {r}_{2}=\frac{\left(4a{\right)}^{2}}{\mathrm{sin}\theta \mathrm{cos}\theta }$
${\left(\frac{{r}_{2}}{{r}_{1}}\right)}^{\frac{1}{3}}=\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }$
$\frac{{\left(\frac{{r}_{2}}{{r}_{1}}\right)}^{\frac{1}{3}}}{{r}_{1}\cdot {r}_{2}}=\frac{{\mathrm{sin}}^{2}\theta }{\left(4a{\right)}^{2}}={r}_{1}^{-\frac{4}{3}}{r}_{1}^{-\frac{2}{3}}$
$\frac{1}{{\left(\frac{{r}_{2}}{{r}_{1}}\right)}^{\frac{1}{3}}{r}_{1}\cdot {r}_{2}}=\frac{{\mathrm{cos}}^{2}\theta }{\left(4a{\right)}^{2}}={r}_{1}^{-\frac{2}{3}}{r}_{1}^{-\frac{4}{3}}$
$\therefore {r}_{1}^{-\frac{4}{3}}{r}_{2}^{-\frac{2}{3}}+{r}_{1}^{-\frac{2}{3}}{r}_{2}^{-\frac{4}{3}}=\frac{1}{\left(4a{\right)}^{2}}$

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