Explanation
{\textbf{Step - 1: Draw the diagram}}{\text{.}}
{\text{At first if we see two given circles , both have a center at (0,0)}}{\text{.}}
{\text{Radius of 1st one, namely }}{{x}^2} + {{y}^2} = 1,{\text{ is 1}}{\text{.}}
{\text{Radius of 2nd one, }}{{x}^2} + {{y}^2} = 4{\text{ , is 2}}{\text{.}}
{\text{Given pair of lines}} \Rightarrow \sqrt 3 \left( {{{x}^2} + {{y}^2}} \right) = 4{xy}{\text{.}}
\Rightarrow \left( {\sqrt 3 } \right){{x}^2} - 4{xy} + \left( {\sqrt 3 } \right){{y}^2} = 0
\sqrt 3 {{x}^2} - 3{xy} - {xy} + + \sqrt 3 {{y}^2} = 0
\Rightarrow \left( {{x} - \sqrt 3 {y}} \right)\left( {\sqrt 3 {x} - {y}} \right).
{\text{So two lines }x} - \sqrt 3 {y} = 0{\text{ and }}\sqrt 3 {x} - {y} = 0.
{\textbf{Step - 2: Find angle between pair of lines}}{\textbf{.}}
{\text{The standard equation of pair of lines is a}}{{x}^2} + 2{hxy} + {b}{{y}^2} = \left( {{x} + {my}} \right)\left( {{px} + {qy}} \right) = 0
{\text{There the angles between them, is }}\theta = {\tan ^{ - 1}}\dfrac{{2\sqrt {{{x}^2} - {ab}} }}{{\left| {\left. {{a} + {b}} \right|} \right.}}
\Rightarrow \theta = {\tan ^{ - 1}}\dfrac{2}{{2\sqrt 3 }}
\Rightarrow \theta = \dfrac{\pi }{6}
{\textbf{Step - 3: Find the area}}{\textbf{.}}
{\text{As per formula area of sectors of two circles is }}\dfrac{\theta }{{2\pi }} \times \pi \left( {{{\text{R}}^2} - {{r}^2}} \right)
{\text{Here R}} = 2,{\text{ and }r} = 1{\text{, }}\theta = \dfrac{\pi }{6}
{\text{So area}} = \dfrac{{\dfrac{\pi }{6}}}{{2\pi }} \times \pi \left( {{2^2} - {1^2}} \right)
= \dfrac{\pi }{{6 \times 2}} \times \left( {4 - 1} \right)
= \dfrac{{\pi \times 3}}{{6 \times 2}}
= \dfrac{\pi }{4}
{\textbf{Thus, the area bounded is }}\boldsymbol{\mathbf{\dfrac{\pi }{4}}{\textbf{ unit}}{^2}}{\text{ }}.
= \cfrac{16}{3}(4 \pi + \sqrt{3})
= (\cfrac{-2x^3}{3} + x^2) | _0 ^1 - \int xlogxdx to Integrate xlogxdx, substitute logx =t, dx= x dtIntegral becomes \int e^{2t}tdt Integration by parts and then putting limits gives = -1/4
The ratio in which the area bounded by the curves y^2=12x and x^2=12y is divided by the line x = 3 is
y^2 =12 x x^2 =12 y \displaystyle \int_0^3 \sqrt{12 x}-\int_0^3 \dfrac{x^2}{12} =\int_0^3 \sqrt{12 x} -\dfrac{x^2}{12} \displaystyle \dfrac{2\sqrt{12}x^{\dfrac{3}{2}}}{3} - \dfrac{x^3}{36}\int_0^3 \displaystyle \dfrac{\sqrt{12}3\sqrt{3}}{3} - \dfrac{27}{36} 12 - \dfrac{3}{4} =\dfrac{45}{4} Same as \dfrac{8}{4} for remaining part \therefore ratio =\dfrac{15}{49}
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