Question

# 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

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 $$r^2$$:
$$\frac{x^2}{9} - \frac{y^2}{16} = 25$$

2021-01-03

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

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

given that - $$\displaystyle{\frac{{{x}^{{2}}}}{{{9}}}}-{\frac{{{y}^{{2}}}}{{{16}}}}={25}$$ put $$x = r \cos 0$$, $$y = e \sin 0$$
$$\Rightarrow \frac{r^2 cos^2 0}{0} - \frac{r^2 sin^2 0}{16} = 25$$
$$\Rightarrow r^2 [\frac{16 cos^2 0 - 9 sin^2 0}{177}] = 25$$
$$\Rightarrow r^2 = \frac{25 \times 177 }{[(7 cos)^2 - (3 sin)^2]} [a^2 - h^2 = (a - b) (a + b)]$$
$$\Rightarrow r^2 = \frac{25 \times 177}{(7 cos 0 - 3 sin 0)(4 cos 0 + 3 sin 0)}$$
$$\Rightarrow r^2 = \frac{25 \times 177}{5[(\frac{7}{5} cos 0 - \frac{3}{5} sin 0)] 5 [(\frac{7}{5} cos 0 + \frac{3}{5} sin 0)]}$$
$$\Rightarrow r^2 \frac{177}{(\frac{7}{5} cos 0 - \frac{3}{5} sin 0)(\frac{7}{5} cos 0 + \frac{3}{5} sin 0)} [sin(53^{\circ}) = \frac{3}{5}]$$
$$\Rightarrow r^2 = \frac{177}{(sin 53^{\circ} cos 0 - cos 53^{\circ} sin 0)(sin 53^{\circ} cos 0 + cos 53^{\circ} sin 0)}$$  formula

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

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

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

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

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