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
{ x = r ⋅ cos θ y = r ⋅ sin θ \left\{\begin{array}{l}
x=r\cdot\cos\theta\\
y=r\cdot\sin\theta
\end{array}\right. { x = r ⋅ cos θ y = r ⋅ sin θ
( More information: https://en.wikipedia.org/wiki/Polar_coordinate_system )
This is enough to solve this task.
(с) y = 2 x y=2x y = 2 x
Then,
r ⋅ sin θ = 2 r ⋅ cos θ → [ r = 0 sin θ = 2 cos θ ∣ ÷ ( cos θ ) [ r = 0 − describes the origin tan θ = 2 − describes other points of the line 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. r ⋅ sin θ = 2 r ⋅ cos θ → [ r = 0 sin θ = 2 cos θ ∣ ÷ ( cos θ ) [ r = 0 − describes the origin tan θ = 2 − describes other points of the line
(d) y = x 2 y=x^2 y = x 2
r ⋅ sin θ = r 2 ⋅ cos 2 θ → [ r = 0 sin θ = r cos 2 θ ∣ ÷ ( cos 2 θ ) [ r = 0 − describes the origin r = sin θ cos 2 θ − describes other points of parabola 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] r ⋅ sin θ = r 2 ⋅ cos 2 θ → [ r = 0 sin θ = r cos 2 θ ∣ ∣ ÷ ( cos 2 θ ) ⎣ ⎡ r = 0 − describes the origin r = cos 2 θ sin θ − describes other points of parabola
(e) x y = 4 xy=4 x y = 4
( r ⋅ sin θ ) ⋅ ( r ⋅ cos θ ) = 4 ∣ × ( 2 ) ⟶ r 2 ⋅ ( 2 sin θ ⋅ cos θ ⏟ s i n 2 θ ) = 8 ⟶ r 2 = 8 sin 2 θ ⟶ r = 8 sin 2 θ \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}}} ( r ⋅ sin θ ) ⋅ ( r ⋅ cos θ ) = 4 ∣ × ( 2 ) ⟶ r 2 ⋅ ( s in 2 θ 2 sin θ ⋅ cos θ ) = 8 ⟶ r 2 = sin 2 θ 8 ⟶ r = sin 2 θ 8
(f) y = 4 y=4 y = 4
r ⋅ sin θ = 4 r\cdot\sin\theta=4 r ⋅ sin θ = 4
(g) 3 x − y + 2 = 0 3x-y+2=0 3 x − y + 2 = 0
3 r ⋅ cos θ − r ⋅ sin θ + 2 = 0 ⟶ r ⋅ ( 3 cos θ − sin θ ) = − 2 ⟶ r = − 2 3 cos θ − sin θ 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} 3 r ⋅ cos θ − r ⋅ sin θ + 2 = 0 ⟶ r ⋅ ( 3 cos θ − sin θ ) = − 2 ⟶ r = 3 cos θ − sin θ − 2
(b) x 2 / a 2 + y 2 / b 2 = 1 x^2/a^2+y^2/b^2=1 x 2 / a 2 + y 2 / b 2 = 1
( r ⋅ cos θ ) 2 a 2 + ( r ⋅ sin θ ) 2 b 2 = 1 ∣ × ( a 2 b 2 ) r 2 ⋅ ( b 2 ⋅ cos 2 θ + a 2 ⋅ sin 2 θ ) = a 2 b 2 r = a 2 b 2 b 2 ⋅ cos 2 θ + a 2 ⋅ sin 2 θ r = a b b 2 ⋅ cos 2 θ + a 2 ⋅ sin 2 θ \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}} a 2 ( r ⋅ cos θ ) 2 + b 2 ( r ⋅ sin θ ) 2 = 1 ∣ ∣ × ( a 2 b 2 ) r 2 ⋅ ( b 2 ⋅ cos 2 θ + a 2 ⋅ sin 2 θ ) = a 2 b 2 r = b 2 ⋅ cos 2 θ + a 2 ⋅ sin 2 θ a 2 b 2 r = b 2 ⋅ cos 2 θ + a 2 ⋅ sin 2 θ ab
Note: for example, I took a = 2 a=2 a = 2 b = 3 b=3 b = 3
(a) ( x − a ) 2 + ( y − a ) 2 = a 2 (x-a)^2+(y-a)^2=a^2 ( x − a ) 2 + ( y − a ) 2 = a 2
( r ⋅ cos θ − a ) 2 + ( r ⋅ sin θ − a ) 2 = a 2 r 2 ⋅ ( cos 2 θ + sin 2 θ ⏟ = 1 ) − 2 a r ⋅ ( cos θ + sin θ ) + 2 a 2 = a 2 r 2 − 2 a r 2 ⋅ ( 2 2 cos θ + 2 2 sin θ ) + a 2 = 0 r 2 − 2 2 a r ⋅ cos ( θ − π 4 ) + a 2 = 0 \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 ( r ⋅ cos θ − a ) 2 + ( r ⋅ sin θ − a ) 2 = a 2 r 2 ⋅ ( = 1 cos 2 θ + sin 2 θ ) − 2 a r ⋅ ( cos θ + sin θ ) + 2 a 2 = a 2 r 2 − 2 a r 2 ⋅ ( 2 2 cos θ + 2 2 sin θ ) + a 2 = 0 r 2 − 2 2 a r ⋅ cos ( θ − 4 π ) + a 2 = 0
Note: for example, I took a = 2 a=2 a = 2
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