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Then we find the enclosed area of two parabolas using an integration formula. Finally we get the required answer.Formula used: The integral formula is \[\int {{{\text{x}}^{\text{n}}}{\text{dx}}} {\text{ = }}\dfrac{{{{\text{x}}^{{\text{n + 1}}}}}}{{{\text{n + 1}}}} + {\text{C}}\], where \[{\text{C}}\] is a constant of integration.Complete step-by-step solution:It is given that that question stated as the equations of the parabolas are \[{{\text{y}}^2}{\text{ = 4ax}}\]………………….\[(1)\]\[{{\text{x}}^2}{\text{ = 4ay}}\]………………….\[(2)\]Here, we have to first find the area of intersection of the two curves: So that we have to find out the point of intersection of the two curves areNow from equation \[(2)\] we can write it as, \[ \Rightarrow \dfrac{{{{\text{x}}^{\text{2}}}}}{{{\text{4a}}}}{\text{ = y}}\]On squaring on both sides,\[ \Rightarrow {\left( {\dfrac{{{{\text{x}}^{\text{2}}}}}{{{\text{4a}}}}} \right)^2}{\text{ = }}{{\text{y}}^2}\]Substitute \[{{\text{y}}^2}\] value,\[ \Rightarrow {\left( {\dfrac{{{{\text{x}}^{\text{2}}}}}{{{\text{4a}}}}} \right)^2}{\text{ = 4ax}}\]\[ \Rightarrow \left( {\dfrac{{{{\text{x}}^4}}}{{{\text{16}}{{\text{a}}^2}}}} \right){\text{ = 4ax}}\]On taking cross multiplying it,\[ \Rightarrow {{\text{x}}^{\text{4}}}{\text{ = 64}}{{\text{a}}^{\text{3}}}{\text{x}}\]On equating this, we get\[ \Rightarrow {{\text{x}}^{\text{4}}} - {\text{64}}{{\text{a}}^{\text{3}}}{\text{x}} = 0\]Taking out the common term,\[ \Rightarrow {\text{x (}}{{\text{x}}^{\text{3}}}{\text{ - 64}}{{\text{a}}^{\text{3}}}{\text{) = 0}}\]Equate to zero, we get\[{\text{x = 0}}\], \[{{\text{x}}^{\text{3}}}{\text{ = 64}}{{\text{a}}^{\text{3}}}\]\[{\text{x = 0}}\], \[{\text{x = 4a}}\]Also, we have to find out \[{\text{y}}\] valuesFrom equation\[(1)\],\[ \Rightarrow \dfrac{{{{\text{y}}^{\text{2}}}}}{{{\text{4a}}}}{\text{ = x}}\]Squaring on both sides,\[ \Rightarrow {\left( {\dfrac{{{{\text{y}}^{\text{2}}}}}{{{\text{4a}}}}} \right)^2}{\text{ = }}{{\text{x}}^2}\]Substitute \[{{\text{x}}^2}\] value,\[ \Rightarrow \dfrac{{{{\text{y}}^{\text{4}}}}}{{{\text{16}}{{\text{a}}^{\text{2}}}}}{\text{ = 4ay}}\] Cross multiplying it,\[ \Rightarrow {{\text{y}}^{\text{4}}}{\text{ = 64}}{{\text{a}}^{\text{3}}}{\text{y}}\] On equating we get\[ \Rightarrow {{\text{y}}^{\text{4}}}{\text{ - 64}}{{\text{a}}^{\text{3}}}{\text{y}} = 0\] Taking out the common term,\[ \Rightarrow {\text{y (}}{{\text{y}}^{\text{3}}}{\text{ - 64}}{{\text{a}}^{\text{3}}}{\text{) = 0}}\]\[{\text{y = 0}}\], \[{{\text{y}}^{\text{3}}}{\text{ = 64}}{{\text{a}}^{\text{3}}}\]\[{\text{y = 0}}\], \[{\text{y = 4a}}\]The point of intersection of these two curves are \[{\text{(0,0)}}\] and \[{\text{(4a,4a)}}\] The required areas is A = (area under parabola\[{{\text{y}}^2}{\text{ = 4ax}}\]) – (area under parabola\[{{\text{x}}^2}{\text{ = 4ay}}\])From equation (1),\[{\text{y = }}\sqrt {{\text{4ax}}} \]From equation (2),\[{\text{y = }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{{\text{4a}}}}\]Here the limit values are\[\left( {0,{\text{ }}4a} \right)\]The enclosed area A = \[\int\limits_{\text{0}}^{{\text{4a}}} {{\text{(}}{{\text{y}}_{\text{2}}}{\text{ - }}{{\text{y}}_{\text{1}}}{\text{)dx}}} \]\[{\text{A = }}\int\limits_{\text{0}}^{{\text{4a}}} {\sqrt {{\text{4ax}}} {\text{dx - }}} \int\limits_{\text{0}}^{{\text{4a}}} {\dfrac{{{{\text{x}}^{\text{2}}}}}{{{\text{4a}}}}{\text{dx}}} \]On splitting the term and we get,\[=\int\limits_{\text{0}}^{{\text{4a}}} {{\text{2}}\sqrt {\text{a}} {{\text{x}}^{\dfrac{{\text{1}}}{{\text{2}}}}}{\text{dx}}} {\text{ - }}\int\limits_{\text{0}}^{{\text{4a}}} {\dfrac{{{{\text{x}}^{\text{2}}}}}{{{\text{4a}}}}{\text{dx}}} \]On integrating this we get, (using integration formula)\[={\text{2}}\sqrt{\text{a}}\left({\dfrac{{{\text{2}}{{\text{x}}^{\dfrac{{\text{3}}}{{\text{2}}}}}}}{{\text{3}}}} \right)_{\text{0}}^{{\text{4a}}}{\text{ - }}\left( {\dfrac{{{{\text{x}}^{\text{3}}}}}{{{\text{12a}}}}} \right)_{\text{0}}^{{\text{4a}}}\]Applying upper and lower limit in the \[{\text{x}}\] term we get,\[=\dfrac{{{\text{4}}\sqrt{\text{a}}}}{{\text{3}}}{\text{(4}}{{\text{a}}^{\dfrac{{\text{3}}}{{\text{2}}}}}{\text{-0) - }}\left( {\dfrac{{{{{\text{(4a)}}}^{\text{3}}}}}{{{\text{12a}}}}} \right)\]After applying limit, we have,\[\Rightarrow\dfrac{{{\text{4}}\sqrt{\text{a}}}}{{\text{3}}}{\text{(8}}{{\text{a}}^{\dfrac{{\text{3}}}{{\text{2}}}}}{\text{) - }}\dfrac{{{\text{64}}{{\text{a}}^{\text{3}}}}}{{1{\text{2a}}}}\]On simplifying this we get,\[=\dfrac{{{\text{32}}}}{{\text{3}}}{{\text{a}}^{\text{2}}}{\text{-}}\dfrac{{{\text{16}}{{\text{a}}^{\text{2}}}}}{{\text{3}}}\]Take out the common term,\[ = \dfrac{{{{\text{a}}^{\text{2}}}}}{{\text{3}}}{\text{(32 - 16)}}\]\[{\text{A = }}\dfrac{{{\text{16}}{{\text{a}}^{\text{2}}}}}{{\text{3}}}\] Square units$\therefore$The enclosed area is \[{\text{ = }}\dfrac{{{\text{16}}{{\text{a}}^{\text{2}}}}}{{\text{3}}}\] square unitsNote: The parabola passes through the origin since \[(0,0)\] satisfies the equation \[{{\text{y}}^2}{\text{ = 4ax}}\]The parabola is the plane curve which is approximately ‘U’ shaped and mirror-symmetrical. The curve \[{{\text{y}}^2}{\text{ = 4ax}}\] is symmetrical about x-axis and hence \[{\text{x}}\]-axis or \[{\text{y = 0}}\] is the axis of the parabola \[{{\text{y}}^2}{\text{ = 4ax}}\]The parabola is symmetric about its axis.Also, the given axis passing through the focus and the vertexRecently Updated PagesMaster Class 11 Business Studies: Engaging Questions & Answers for SuccessMaster Class 11 Computer Science: Engaging Questions & Answers for SuccessMaster Class 11 Economics: Engaging Questions & Answers for SuccessMaster Class 11 Social Science: Engaging Questions & Answers for SuccessMaster Class 11 English: Engaging Questions & Answers for SuccessMaster Class 11 Chemistry: Engaging Questions & Answers for SuccessMaster Class 11 Business Studies: Engaging Questions & Answers for SuccessMaster Class 11 Computer Science: Engaging Questions & Answers for SuccessMaster Class 11 Economics: Engaging Questions & Answers for SuccessMaster Class 11 Social Science: Engaging Questions & Answers for SuccessMaster Class 11 English: Engaging Questions & Answers for SuccessMaster Class 11 Chemistry: Engaging Questions & Answers for Success

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