Answer to Question #111793 in Analytic Geometry for Kwesi Tsiks

Question

Find the polar equations of the lines or curves whose Cartesian equations are
(a) (x x a)
2 + (y y a)
2 = a
2
(b) x
2/a2 + y
2/b2 = 1 (c) y = 2x (d) y = x
2
(e) xy = 4 (f) y = 4 (g) 3x x y + 2 = 0.

Answer 1

Note: for verification I will use a graphical online calculator -https://www.desmos.com/calculator

As we know the Cartesian coordinates are related to polar related relationships

$$\left\{\begin{array}{l} x=r\cdot\cos\theta\\ y=r\cdot\sin\theta \end{array}\right.$$

( More information: https://en.wikipedia.org/wiki/Polar_coordinate_system )

This is enough to solve this task.

(с) $y=2x$

Then,

$$r\cdot\sin\theta=2r\cdot\cos\theta\to \left[\begin{array}{l} r=0\\ \left.\sin\theta=2\cos\theta\right|\div\left(\cos\theta\right) \end{array}\right.\\[0.3cm] \left[\begin{array}{l} r=0-\text{describes the origin}\\ \tan\theta=2-\text{describes other points of the line} \end{array}\right.$$

(d) $y=x^2$

$$r\cdot\sin\theta=r^2\cdot\cos^2\theta\to \left[\begin{array}{l} r=0\\ \left.\sin\theta=r\cos^2\theta\right|\div\left(\cos^2\theta\right) \end{array}\right.\\[0.3cm] \left[\begin{array}{l} r=0-\text{describes the origin}\\[0.3cm] r=\displaystyle\frac{\sin\theta}{\cos^2\theta}-\text{describes other points of parabola} \end{array}\right.\\[0.3cm]$$

(e) $xy=4$

$$\left.\left(r\cdot\sin\theta\right)\cdot\left(r\cdot\cos\theta\right)=4\right|\times(2)\longrightarrow\\[0.3cm] r^2\cdot\left(\underbrace{2\sin\theta\cdot\cos\theta}_{sin2\theta}\right)=8\longrightarrow\\[0.3cm] r^2=\frac{8}{\sin2\theta}\longrightarrow\boxed{r=\sqrt{\frac{8}{\sin2\theta}}}$$

(f) $y=4$

$$r\cdot\sin\theta=4$$

(g) $3x-y+2=0$

$$3r\cdot\cos\theta-r\cdot\sin\theta+2=0\longrightarrow\\[0.3cm] r\cdot\left(3\cos\theta-\sin\theta\right)=-2\longrightarrow\\[0.3cm] r=\frac{-2}{3\cos\theta-\sin\theta}$$

(b) $x^2/a^2+y^2/b^2=1$

$$\left.\frac{\left(r\cdot\cos\theta\right)^2}{a^2}+\frac{\left(r\cdot\sin\theta\right)^2}{b^2}=1\right|\times\left(a^2b^2\right)\\[0.3cm] r^2\cdot\left(b^2\cdot\cos^2\theta+a^2\cdot\sin^2\theta\right)=a^2b^2\\[0.3cm] r=\sqrt{\frac{a^2b^2}{b^2\cdot\cos^2\theta+a^2\cdot\sin^2\theta}}\\[0.3cm] r=\frac{ab}{\sqrt{b^2\cdot\cos^2\theta+a^2\cdot\sin^2\theta}}$$

Note: for example, I took $a=2$ and $b=3$

(a) $(x-a)^2+(y-a)^2=a^2$

$$\left(r\cdot\cos\theta-a\right)^2+\left(r\cdot\sin\theta-a\right)^2=a^2\\[0.3cm] r^2\cdot\left(\underbrace{\cos^2\theta+\sin^2\theta}_{=1}\right)-2ar\cdot\left(\cos\theta+\sin\theta\right)+2a^2=a^2\\[0.3cm] r^2-2ar\sqrt{2}\cdot\left(\frac{\sqrt{2}}{2}\cos\theta+\frac{\sqrt{2}}{2}\sin\theta\right)+a^2=0\\[0.3cm] r^2-2\sqrt{2}ar\cdot\cos\left(\theta-\frac{\pi}{4}\right)+a^2=0$$

Note: for example, I took $a=2$