# 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

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

2021-01-03
Solution: Relation between polar Co-ordinate system and Cartesian Co-ordinate system $$\displaystyle{x}={r}{\cos{{0}}}{N}{S}{K}{y}={r}{\sin{{0}}}{N}{S}{K}{r}=\sqrt{{{x}^{{2}}+{y}^{{2}}}}$$ given that - $$\displaystyle{\frac{{{x}^{{2}}}}{{{9}}}}-{\frac{{{y}^{{2}}}}{{{16}}}}={25}$$ put x = r cos 0, y = e sin 0
\Rightarrow \frac{r^2 cos^2 0}{0} - \frac{r^2 sin^2 0}{16} = 25
\Rightarrow r^2 [\frac{16 cos^2 0 - 9 sin^2 0}{177}] = 25
\Rightarrow r^2 = \frac{25 \times 177 }{[(7 cos)^2 - (3 sin)^2]} [a^2 - h^2 = (a - b) (a + b)]
\Rightarrow r^2 = \frac{25 \times 177}{(7 cos 0 - 3 sin 0)(4 cos 0 + 3 sin 0)}
\Rightarrow r^2 = \frac{25 \times 177}{5[(\frac{7}{5} cos 0 - \frac{3}{5} sin 0)] 5 [(\frac{7}{5} cos 0 + \frac{3}{5} sin 0)]}
\Rightarrow r^2 \frac{177}{(\frac{7}{5} cos 0 - \frac{3}{5} sin 0)(\frac{7}{5} cos 0 + \frac{3}{5} sin 0)} [sin(53^{\circ}) = \frac{3}{5}]
\Rightarrow r^2 = \frac{177}{(sin 53^{\circ} cos 0 - cos 53^{\circ} sin 0)(sin 53^{\circ} cos 0 + cos 53^{\circ} sin 0)}ZSK formula $$\displaystyle{\sin{{\left({A}+{B}\right)}}}={\sin{{A}}}{\cos{{B}}}+{\cos{{A}}}{\sin{{B}}}{N}{S}{K}{\sin{{\left({A}-{B}\right)}}}={\sin{{A}}}{\cos{{B}}}-{\cos{{A}}}{\sin{{B}}}{N}{S}{K}\Rightarrow{r}^{{2}}{\frac{{{177}}}{{{\left[{\sin{{\left({53}-{0}\right)}}}\right]}{\left[{\sin{{\left({53}+{0}\right)}}}\right]}}}}{N}{S}{K}\Rightarrow{r}^{{2}}={\frac{{{177}}}{{{\left[{\sin{{\left({53}-{0}\right)}}}\right]}{\left[{\sin{{\left({0}+{53}\right)}}}\right]}}}}$$ Finally answer is $$\displaystyle\Rightarrow{r}^{{2}}={\frac{{{177}}}{{{\left[{\sin{{\left({53}-{0}\right)}}}\right]}{\left[{\sin{{\left({0}+{53}\right)}}}\right]}}}}$$

### Relevant Questions

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Given:
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$$\displaystyle\ {y}=\ {r}{\sin{\theta}}$$
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$$\displaystyle{r}=\pm\sqrt{{{x}^{{{2}}}\ +\ {y}^{{{2}}}}}$$
$$\displaystyle{\cos{\theta}}={\frac{{{x}}}{{{r}}}},{\sin{\theta}}={\frac{{{y}}}{{{r}}}},{\tan{\theta}}={\frac{{{x}}}{{{y}}}}$$
Calculation:
Given: equation in rectangular-coordinate is $$\displaystyle{y}={x}$$.
Converting into equivalent polar equation -
$$\displaystyle{y}={x}$$
Put $$\displaystyle{x}={r}{\cos{\theta}},\ {y}={r}{\sin{\theta}},$$
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