# Solve the given Alternate Coordinate Systems and give a correct answer10) Convert the equation from Cartesian to polar coordinates solving for r^2: frac{x^2}{9} - frac{y^2}{16} = 25

Solve the given Alternate Coordinate Systems and give a correct answer 10) Convert the equation from Cartesian to polar coordinates solving for ${r}^{2}$:
$\frac{{x}^{2}}{9}-\frac{{y}^{2}}{16}=25$

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Solution: Relation between polar Co-ordinate system and Cartesian Co-ordinate system $x=r\mathrm{cos}0y=r\mathrm{sin}0$

$r=\sqrt{{x}^{2}+{y}^{2}}$

given that - $\frac{{x}^{2}}{9}-\frac{{y}^{2}}{16}=25$ put $x=r\mathrm{cos}0$, $y=e\mathrm{sin}0$
$⇒\frac{{r}^{2}co{s}^{2}0}{0}-\frac{{r}^{2}si{n}^{2}0}{16}=25$
$⇒{r}^{2}\left[\frac{16co{s}^{2}0-9si{n}^{2}0}{177}\right]=25$
$⇒{r}^{2}=\frac{25×177}{\left[\left(7cos{\right)}^{2}-\left(3sin{\right)}^{2}\right]}\left[{a}^{2}-{h}^{2}=\left(a-b\right)\left(a+b\right)\right]$
$⇒{r}^{2}=\frac{25×177}{\left(7cos0-3sin0\right)\left(4cos0+3sin0\right)}$
$⇒{r}^{2}=\frac{25×177}{5\left[\left(\frac{7}{5}cos0-\frac{3}{5}sin0\right)\right]5\left[\left(\frac{7}{5}cos0+\frac{3}{5}sin0\right)\right]}$
$⇒{r}^{2}\frac{177}{\left(\frac{7}{5}cos0-\frac{3}{5}sin0\right)\left(\frac{7}{5}cos0+\frac{3}{5}sin0\right)}\left[sin\left({53}^{\circ }\right)=\frac{3}{5}\right]$
$⇒{r}^{2}=\frac{177}{\left(sin{53}^{\circ }cos0-cos{53}^{\circ }sin0\right)\left(sin{53}^{\circ }cos0+cos{53}^{\circ }sin0\right)}$  formula

$\mathrm{sin}\left(A+B\right)=\mathrm{sin}A\mathrm{cos}B+\mathrm{cos}A\mathrm{sin}B$

$\mathrm{sin}\left(A-B\right)=\mathrm{sin}A\mathrm{cos}B-\mathrm{cos}A\mathrm{sin}B$

$⇒{r}^{2}\frac{177}{\left[\mathrm{sin}\left(53-0\right)\right]\left[\mathrm{sin}\left(53+0\right)\right]}$

$⇒{r}^{2}=\frac{177}{\left[\mathrm{sin}\left(53-0\right)\right]\left[\mathrm{sin}\left(0+53\right)\right]}$

Finally answer is $⇒{r}^{2}=\frac{177}{\left[\mathrm{sin}\left(53-0\right)\right]\left[\mathrm{sin}\left(0+53\right)\right]}$