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This is one of our many quizzes on Polar and rectangular equations. Click on this to access more such competitive quizzes on 2D Coordinate Geometry.

Polar and rectangular equations Quiz 2

mathematics

 Convert  $2\theta =\dfrac{\pi }{2}$  to rectangular form.

 Substitute $\theta ={{\tan }^{-1}}\left( \dfrac{Y}{X} \right)$ and then simplify.

  $\text{Given; } 2\theta =\dfrac{\pi }{2} \\ \Rightarrow \theta =\dfrac{\pi }{4} \\ \Rightarrow {{\tan }^{-1}}\left( \dfrac{Y}{X} \right)=\dfrac{\pi }{4} \\ \Rightarrow \left( \dfrac{Y}{X} \right)=\tan \left( \dfrac{\pi }{4} \right)=1 \\ \therefore X=Y$

 What is the difference between the rectangular form of  $4R=\sqrt{2}\cos \theta $ and $3R=5\sin \theta $.

 Substitute $X=R\cos \theta ,Y=R\sin \theta ,{{X}^{2}}+{{Y}^{2}}={{R}^{2}}$ to get the rectangular equation. Find the required equations separately and then subtract them to get the answer.

  The given polar form is, $\\ 4R=\sqrt{2}\cos \theta  \\ \Rightarrow 4R=\sqrt{2}\left( \dfrac{X}{R} \right) \\ \therefore 4{{R}^{2}}=\sqrt{2}X \\ \Rightarrow 4{{X}^{2}}+4{{Y}^{2}}-\sqrt{2}X=0..........(I) \\ $ The other polar equation is, $\\ 3R=5\sin \theta  \\ \Rightarrow 3R=5\left( \dfrac{Y}{R} \right) \\ \Rightarrow 3{{X}^{2}}+3{{Y}^{2}}-5Y=0..........(II) \\ $ Subtracting equation (I) and (II), we get $\\ 4{{X}^{2}}+4{{Y}^{2}}-\sqrt{2}X-\left( 3{{X}^{2}}+3{{Y}^{2}}-5Y \right)=0 \\ \Rightarrow {{X}^{2}}+{{Y}^{2}}-\sqrt{2}X+5Y=0 $

 Find the sum of rectangular equations of $R=4\sin \theta +5\cos \theta $ and $R=\dfrac{3}{4\sin \theta +5\cos \theta }$.

 Substitute $X=R\cos \theta ,Y=R\sin \theta ,{{X}^{2}}+{{Y}^{2}}={{R}^{2}}$ to get the rectangular equation. Find the required equations separately and then add them to get the answer.

  Given, $\\ R=4\sin \theta +5\cos \theta  \\ \Rightarrow R=4\left( \dfrac{Y}{R} \right)+5\left( \dfrac{X}{R} \right) \\ \therefore {{R}^{2}}=4Y+5X \\ \Rightarrow {{X}^{2}}+{{Y}^{2}}=4Y+5X..........(I) \\ $ The other polar equation is, $\\ R=\dfrac{3}{4\sin \theta +5\cos \theta } \\ \Rightarrow R=\dfrac{3}{4\left( \dfrac{Y}{R} \right)+5\left( \dfrac{X}{R} \right)} \\ \Rightarrow R=\dfrac{3R}{4Y+5X} \\ \Rightarrow 4Y+5X=3..........(II) \\ $ Adding equation (I) and (II), we get $\\ {{X}^{2}}+{{Y}^{2}}+4Y+5X=3+4Y+5X \\ \Rightarrow {{X}^{2}}+{{Y}^{2}}=3 $

 Find the product of polar forms of $3X+4Y=4$ and reciprocal of $6X+8Y=12$

 Substitute $X=R\cos \theta ,Y=R\sin \theta ,{{X}^{2}}+{{Y}^{2}}={{R}^{2}}$ to get the polar equation. Find the required equations separately and then multiply the first equation with the reciprocal of the second equation to get the answer.

  The given rectangular equation is, $\\ 3X+4Y=4 \\ \Rightarrow 3\left( R\cos \theta  \right)+4\left( R\sin \theta  \right)=4 \\ \Rightarrow 3\cos \theta +4\sin \theta =\dfrac{4}{R}..........(I) \\ $ The other rectangular equation is, $\\ 6X+8Y=12 \\ \Rightarrow 6\left( R\cos \theta  \right)+8\left( R\sin \theta  \right)=12 \\ \Rightarrow 6\cos \theta +8\sin \theta =\dfrac{12}{R}..........(II) \\ $ The product of polar forms of $3X+4Y=4$ and reciprocal of $6X+8Y=12$ from equation (I) and (II), $\\ \left( 3\cos \theta +4\sin \theta  \right)\times \left( \dfrac{1}{6\cos \theta +8\sin \theta } \right)=\dfrac{4}{R}\times \dfrac{R}{12} \\ \Rightarrow \dfrac{1}{3} \\ $

 What is the ratio of polar equations of  $3{{Y}^{2}}=2X$ and $9{{X}^{2}}=8Y$ when $\theta =\dfrac{\pi }{3}$.

 Substitute $X=R\cos \theta ,Y=R\sin \theta ,{{X}^{2}}+{{Y}^{2}}={{R}^{2}}$ to get the polar equation. Find the required equations separately and then find the ratio by dividing the polar equations.

  The given rectangular equation is, $\\ 3{{Y}^{2}}=2X \\ \Rightarrow 3\left( {{R}^{2}}{{\sin }^{2}}\theta  \right)=2\left( R\cos \theta  \right) \\ \Rightarrow 3R{{\sin }^{2}}\theta =2\cos \theta  \\ \text{When }\theta =\dfrac{\pi }{3}, \\ 3R{{\sin }^{2}}\left( \dfrac{\pi }{3} \right)=2\cos \left( \dfrac{\pi }{3} \right) \\ \Rightarrow \dfrac{9R}{4}=1........(i) \\ $ The other rectangular equation is, $\\ 9{{X}^{2}}=8Y \\ \Rightarrow 9\left( {{R}^{2}}{{\cos }^{2}}\theta  \right)=8\left( R\sin \theta  \right) \\ \Rightarrow 9R{{\cos }^{2}}\theta =8\sin \theta  \\ \text{When }\theta =\dfrac{\pi }{3}, \\ 9R{{\cos }^{2}}\left( \dfrac{\pi }{3} \right)=8\sin \left( \dfrac{\pi }{3} \right) \\ \Rightarrow \dfrac{9R}{4}=4\sqrt{3}........(ii) \\ $ Required ratio is,$\\ \dfrac{\dfrac{9R}{4}}{\dfrac{9R}{4}}=\dfrac{1}{4\sqrt{3}}$