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This is one of our many quizzes on Polar coordinates and parametric equations. Click on this to access more such competitive quizzes on Conic Sections.

Polar coordinates and parametric equations Quiz 2

mathematics

 If $x = {e^{\sin \left( {\dfrac{{2m + 1}}{3}} \right)}}$ and $y = {e^{\cos \left( {\dfrac{{2m + 1}}{3}} \right)}}$ then the value of $\dfrac{y}{x}$ at $m =  - \dfrac{1}{2}$ is

 Simplify $\dfrac{y}{x}$ and substitute $m =  - \dfrac{1}{2}$.

Given $x = {e^{\sin \left( {\dfrac{{2m + 1}}{3}} \right)}}$ and $y = {e^{\cos \left( {\dfrac{{2m + 1}}{3}} \right)}} \\ \Rightarrow \dfrac{y}{x} = \dfrac{{{e^{\cos \left( {\dfrac{{2m + 1}}{3}} \right)}}}}{{{e^{\sin \left( {\dfrac{{2m + 1}}{3}} \right)}}}} = {e^{\cos \left( {\dfrac{{2m + 1}}{3}} \right) - \sin \left( {\dfrac{{2m + 1}}{3}} \right)}} \\ $ Put $m =  - \dfrac{1}{2}$ $ \\ \Rightarrow \dfrac{y}{x} = {e^{\cos 0 - \sin 0}} = {e^1} = e$

 If $x = \log \left( {\dfrac{{19}}{t}} \right)$ and $y = \log \left( {\dfrac{{76}}{t}} \right)$, for $t > 0$, then the value of $x + y$ at $t = 19$ is

 Simplify $x + y$ and substitute $t = 19$.

$x + y = \log \left( {\dfrac{{19}}{t}} \right) + \log \left( {\dfrac{{76}}{t}} \right) \\   \Rightarrow x + y = \log \left( {\dfrac{{19 \times 76}}{{{t^2}}}} \right) \\  $ Put $t = 19$. $ \\ \Rightarrow x + y = \log \left( {\dfrac{{19 \times 76}}{{19 \times 19}}} \right) = \log \left( 4 \right) = \log {2^2} \\   \Rightarrow x + y = 2\log 2 \\  $

 The equation $r = 2\left( {\cos \theta  + \sin \theta } \right)$ written in cartesian form is

 Substitute $x = r\cos \theta $, $y = r\sin \theta $ and ${r^2} = {x^2} + {y^2}$.

Given $r = 2\left( {\cos \theta  + \sin \theta } \right)$ $ \\ $ Put $x = r\cos \theta $, $y = r\sin \theta $ and ${r^2} = {x^2} + {y^2}$. $ \\  \Rightarrow r = 2\left( {\dfrac{x}{r} + \dfrac{y}{r}} \right) \\   \Rightarrow {r^2} = 2x + 2y \Rightarrow {x^2} + {y^2} = 2x + 2y \\   \Rightarrow {x^2} - 2x + {y^2} - 2y = 0 \\   \Rightarrow {x^2} - 2x + 1 + {y^2} - 2y + 1 - 2 = 0 \\   \Rightarrow {\left( {x - 1} \right)^2} + {\left( {y - 1} \right)^2} = 2 \\  $ 

 The equation of an ellipse is given as $r = \dfrac{9}{{4 + 2\cos \theta }}$. The semi-major and semi-minor axis are

 Convert the given equation into Cartesian form and compare with the standard form of the equation to get the semi-major and semi-minor axis.

$ r = \dfrac{8}{{5 + 3\cos \theta }} \\   \Rightarrow 5r + 3r\cos \theta  = 8 \\   \Rightarrow 5r = 8 - 3r\cos \theta  \\  $ Square both sides and put $x = r\cos \theta ,\;{r^2} = {x^2} + {y^2}$. $ \\  \Rightarrow {\left( {5r} \right)^2} = {\left( {8 - 3x} \right)^2} \\   \Rightarrow 25{r^2} = 81 + 9{x^2} - 48x \\   \Rightarrow 25\left( {{x^2} + {y^2}} \right) = 81 + 9{x^2} - 48x \Rightarrow 25{x^2} + 25{y^2} = 81 + 9{x^2} - 48x \\   \Rightarrow 16{x^2} - 48x + 25{y^2} = 81 \\  $ This can be simplified as, $ \\  \Rightarrow 16{x^2} - 48x + 25{y^2} = 81 \\   \Rightarrow {\left( {4x} \right)^2} - 2 \times 4x \times 6 + {6^2} + {\left( {5y} \right)^2} = 81 - 36 \\   \Rightarrow {\left( {4x - 6} \right)^2} + {\left( {5y} \right)^2} = 45 \\   \Rightarrow \dfrac{{{{\left( {4x - 6} \right)}^2}}}{{45}} + \dfrac{{{{\left( {5y} \right)}^2}}}{{45}} = 1 \\   \Rightarrow \dfrac{{{{\left( {x - \dfrac{3}{2}} \right)}^2}}}{{{{\left( {\sqrt {\dfrac{{45}}{{16}}} } \right)}^2}}} + \dfrac{{{y^2}}}{{{{\left( {\sqrt {\dfrac{{45}}{{25}}} } \right)}^2}}} = 1 \\  $ On comparing with standard form of equation $\dfrac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} + \dfrac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1$ $ \\ $ Semi-major axis is $a = \sqrt {\dfrac{{45}}{{16}}}  = \dfrac{{3\sqrt 5 }}{4}$ $ \\ $ And semi-minor axis is $b = \sqrt {\dfrac{{45}}{{25}}}  = \dfrac{{3\sqrt 5 }}{5}$

 If $x = \sqrt {\dfrac{{12 + 12\cos \left( {\dfrac{{17u}}{5}} \right)}}{{12 - 12\cos \left( {\dfrac{{17u}}{5}} \right)}}} $ and $y = \cot \left( {\dfrac{{17u}}{{10}}} \right)$, then the value of ${\left( {\dfrac{y}{x}} \right)^2}$ at $u = 5$ is

 Simplify $x$ and then evaluate $\dfrac{y}{x}$.

$ x = \sqrt {\dfrac{{12 + 12\cos \left( {\dfrac{{17u}}{5}} \right)}}{{12 - 12\cos \left( {\dfrac{{17u}}{5}} \right)}}}  \\   \Rightarrow x = \sqrt {\dfrac{{12\left( {1 + \cos \left( {\dfrac{{17u}}{5}} \right)} \right)}}{{12\left( {1 - \cos \left( {\dfrac{{17u}}{5}} \right)} \right)}}}  \\   \Rightarrow x = \sqrt {\dfrac{{2{{\cos }^2}\left( {\dfrac{{17u}}{{10}}} \right)}}{{2{{\sin }^2}\left( {\dfrac{{17u}}{{10}}} \right)}}}  = \sqrt {{{\cot }^2}\left( {\dfrac{{17u}}{{10}}} \right)}  = \cot \left( {\dfrac{{17u}}{{10}}} \right) \\   \Rightarrow \dfrac{y}{x} = \dfrac{{\cot \left( {\dfrac{{17u}}{{10}}} \right)}}{{\cot \left( {\dfrac{{17u}}{{10}}} \right)}} = 1 \\   \Rightarrow {\left( {\dfrac{y}{x}} \right)^2} = 1 \\  $