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This is one of our many quizzes on Rectangular and polar coordinates. Click on this to access more such competitive quizzes on Number System.

Rectangular and polar coordinates Quiz 2

mathematics

 The trajectory of the missile on the polar coordinate system is traced and at any point $\left( r,\theta  \right)$ trajectory is $\dfrac{2}{r}=\sin \theta -\sec \theta $. Then, what is the trajectory of the missile in Cartesian form?

 Using the relation between the polar coordinates and rectangular coordinates, obtain the Cartesian equation for trajectory of the missile.

\[\begin{align} &nbsp;&nbsp;& \text{Cartesian coordinates }\left( x,y \right)\text{ can be converted into polar coordinates }\left( r,\theta \right)\text{ by using formula,} \\ &nbsp;& x=r\cos \theta \text{ and }y=r\sin \theta ,\text{ where }r=\sqrt{{{x}^{2}}+{{y}^{2}}}. \\ &nbsp;& \text{So, the trajectory of missile }\dfrac{2}{r}=\sin \theta -\sec \theta \text{ in cartesian form is,} \\ &nbsp;& \Rightarrow \dfrac{2}{r}=\sin \theta -\sec \theta \\ &nbsp;& \Rightarrow 2\cos \theta =r\sin \theta \cos \theta -r \\ &nbsp;& \Rightarrow 2r\cos \theta ={{r}^{2}}\sin \theta \cos \theta -{{r}^{2}} \\ &nbsp;& \Rightarrow 2x=xy-\left( {{x}^{2}}+{{y}^{2}} \right) \\ \end{align}\]

 At what coordinate does the center of $r=2\cos \theta $ lie in the rectangular coordinate system?

 Convert polar equation into rectangular coordinate equation and use completing the square method to evaluate the center of the circle.

$\begin{align} &nbsp;&nbsp;& \text{If }\left( x,y \right)\text{ are cartesian coordinates, then in polar form }\left( r,\theta \right)\text{ it is,} \\ &nbsp;& x=r\cos \theta \text{ and }y=r\sin \theta ,\text{ where r}=\sqrt{{{x}^{2}}+{{y}^{2}}}. \\ &nbsp;& \text{So,} \\ &nbsp;& \Rightarrow r=2\cos \theta \\ &nbsp;& \Rightarrow {{r}^{2}}=2r\cos \theta \\ &nbsp;& \Rightarrow {{x}^{2}}+{{y}^{2}}=2x \\ &nbsp;& \Rightarrow {{\left( x-1 \right)}^{2}}+{{y}^{2}}=1 \\ &nbsp;& \text{So, center of circle is (1,0)}\text{.} \\ \end{align}$

 If polar representation of point $\left( \sqrt{3},z \right)$is$\left( z+1,\dfrac{\pi }{6} \right)$, then value of ${{\left( z+2 \right)}^{2}}-9$ is equal to

 Change the Cartesian coordinate into polar coordinate and obtain the value of ‘z’, and then solve for the value of ${{\left( z+2 \right)}^{2}}-9$.

\[\begin{align} &nbsp;&nbsp;& \text{If }\left( x,y \right)\text{ are cartesian coordinates, then in polar form }\left( r,\theta \right)\text{ it is,} \\ &nbsp;& x=r\cos \theta \text{ and }y=r\sin \theta ,\text{ where r}=\sqrt{{{x}^{2}}+{{y}^{2}}}. \\ &nbsp;& \text{So, } \\ &nbsp;& \Rightarrow \sqrt{3}=\left( z+1 \right)\cos \left( \dfrac{\pi }{6} \right) \\ &nbsp;& \Rightarrow 2=z+1 \\ &nbsp;& \Rightarrow z=1 \\ &nbsp;& \text{Then, } \\ &nbsp;& \Rightarrow {{\left( z+2 \right)}^{2}}-9 \\ &nbsp;& \Rightarrow {{\left( 1+2 \right)}^{2}}-9 \\ &nbsp;& \Rightarrow 0 \\ \end{align}\]

 If line $r=\dfrac{1}{\cos \theta -\sin \theta }$ in rectangular coordinate system intersects line $x+y=1$, then the value of $\left( x,y \right)$ is

 Convert polar line into rectangular coordinate line and solve both equations for intersection point.

$\begin{align} &nbsp;&nbsp;& \text{If }\left( x,y \right)\text{ are cartesian coordinates, then in polar form }\left( r,\theta \right)\text{ it is,} \\ &nbsp;& x=r\cos \theta \text{ and }y=r\sin \theta ,\text{ where r}=\sqrt{{{x}^{2}}+{{y}^{2}}}. \\ &nbsp;& \text{So}, \\ &nbsp;& \Rightarrow r=\dfrac{1}{\cos \theta -\sin \theta } \\ &nbsp;& \Rightarrow r\cos \theta -r\sin \theta =1 \\ &nbsp;& \Rightarrow x-y=1 \\ &nbsp;& \text{Let, }x-y=1\text{ intersect at point }\left( a,b \right),\text{ then} \\ &nbsp;& \text{at }\left( a,b \right), \\ &nbsp;& \Rightarrow x-y=x+y \\ &nbsp;& \Rightarrow y=0 \\ &nbsp;& \Rightarrow x=1 \\ &nbsp;& \text{Thus, point of intersection is }\left( 1,0 \right). \\ \end{align}$

 When the road joining the houses at the points $\left( 2\sqrt{10},{{\tan }^{-1}}(-3) \right)$ and $\left( 8,{{\tan }^{-1}}\left( 2 \right) \right)$ was mapped on the rectangular plotting paper, it was seen that road was intersecting another road joining the houses at the point $\left( 8,12 \right)$ and $\left( x,24 \right)$ at ${{90}^{{}^\circ }}$. If, $\cos \left( {{\tan }^{-1}}2 \right)=\dfrac{1}{2}\text{ and }\sin \left( {{\tan }^{-1}}2 \right)=1$ and $\cos \left( {{\tan }^{-1}}\left( -3 \right) \right)=-\dfrac{1}{\sqrt{10}}\text{ and }\sin \left( {{\tan }^{-1}}\left( -3 \right) \right)=\dfrac{3}{\sqrt{10}}$ then, what is the value of $x$?

 Use the conversion of polar coordinates into rectangular coordinates, then find the slopes of roads and obtain the value of $x$ by considering the condition of two perpendicular lines.

$\begin{align} &nbsp;&nbsp;& \text{Polar coordinate }\left( r,\theta \right)\text{ in cartesian plane is represented as,} \\ &nbsp;& x=r\cos \theta \text{ and y}=rsin\theta ,\text{ where }r=\sqrt{{{x}^{2}}+{{y}^{2}}}. \\ &nbsp;& \text{So, point }\left( 2\sqrt{10},{{\tan }^{-1}}(-3) \right)\text{ in rectangular coordinate is,} \\ &nbsp;& \Rightarrow \left( 2\sqrt{10}\cos \left( {{\tan }^{-1}}(-3) \right),2\sqrt{10}sin\left( {{\tan }^{-1}}(-3) \right) \right) \\ &nbsp;& \Rightarrow \left( -2,6 \right) \\ &nbsp;& \text{and, point }\left( 8,{{\tan }^{-1}}\left( 2 \right) \right)\text{ in rectangular coordinate is,} \\ &nbsp;& \Rightarrow \left( 8\cos \left( {{\tan }^{-1}}\left( 2 \right) \right),8sin\left( {{\tan }^{-1}}(2) \right) \right) \\ &nbsp;& \Rightarrow \left( 4,8 \right) \\ &nbsp;& \text{Slope of line with points }\left( a,b \right)\text{ and }\left( c,d \right)\text{ on it is, }\dfrac{d-b}{c-a}. \\ &nbsp;& \text{So, the slope of road joining houses at }\left( -2,6 \right)\text{ and }\left( 4,8 \right)\text{ is,} \\ &nbsp;& \Rightarrow \dfrac{8-6}{4-(-2)}=\dfrac{1}{3} \\ &nbsp;& \text{and, the slope of road joining houses at }\left( 8,12 \right)\text{ and }\left( x,24 \right)\text{ is,} \\ &nbsp;& \Rightarrow \dfrac{24-12}{x-8}=\dfrac{12}{x-8} \\ &nbsp;& \text{Product of slope of two perpendicular lines is equal to -1}\text{.} \\ &nbsp;& \Rightarrow \dfrac{12}{x-8}\times \dfrac{1}{3}=-1 \\ &nbsp;& \Rightarrow x=4 \\ \end{align}$