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  • Find the area of the region bounded by the parabola y2 = 2px, x2 = 2py

Find the area of the region bounded by the parabola y^2 = 2px, x^2 = 2py

Answers (1)

The equation of the parabola is y\textsuperscript{2 }= 2px which shows no negative values for x. Therefore, the graph will be on right of Y-axis will be passing through (0, 0)

Similarly, for equations x\textsuperscript{2} = 2py no negative values for y, therefore, the graph will be above the X axis passing through (0, 0).

For finding point of interaction, let us solve the equations simultaneously. 

\\ \text { Put } y=\frac{x^{2}}{2 p} \text { in } y^{2}=2 p x \\ \Rightarrow\left(\frac{x^{2}}{2 p}\right)^{2}=2 p x \\ \Rightarrow \frac{x^{4}}{4 p^{2}}=2 p x \\ \Rightarrow x^{4}=8 p^{3} x \\ \Rightarrow x^{3}=8 p^{3} \\

\\ \Rightarrow x=2 p \\ \text { Put } x=2 p \text { in } y^{2}=2 p x \\ \Rightarrow y^{2}=2 p(2 p) \\ \Rightarrow y=2 p

The interaction point was found to be (2p, 2p)

Now, we have to find out the areas between the two parabolas.

The equation can be written as

 Area between the two parabola = area under parabola 

y\textsuperscript{2 }= 2px - \text{area under the parabola } x\textsuperscript{2 }= 2py $ \ldots $ (1) \\

For finding the area undery\textsuperscript{2 }= 2px\: \: parabola \\

 

            \item y = \sqrt 2p \sqrt x \\

 

Integrating the equation from 0 to 2p 

\\ \Rightarrow \int_{0}^{2 p} y d x=\sqrt{2 p} \int_{0}^{2 p} \sqrt{x} d x \\ \Rightarrow \int_{0}^{2 p} y d x=\sqrt{2 p} \int_{0}^{2 p} x^{\frac{1}{2}} d x \\ \Rightarrow \int_{0}^{2 p} y d x=\sqrt{2 p}\left[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{2 p} \\ \Rightarrow \int_{0}^{2 p} y d x=\sqrt{2 p}\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{2 p} \\ \Rightarrow \int_{0}^{2 p} y d x=\sqrt{2 p} \frac{2}{3}\left[(2 p)^{\frac{3}{2}}-0\right]

\\\Rightarrow \int_{0}^{2 p} y d x=\frac{2}{3}(2 p)^{\frac{1}{2}}(2 p)^{\frac{3}{2}} \\ \Rightarrow \int_{0}^{2 p} y d x=\frac{2}{3}(2 p)^{\frac{1}{2}+\frac{3}{2}} \\ \Rightarrow \int_{0}^{2 p} y d x=\frac{2}{3}(2 p)^{2} \\ \Rightarrow \int_{0}^{2 p} y d x=\frac{8 p^{2}}{3}

For finding the area under x\textsuperscript{2 }= 2py \text{ parabola} \\

 

            \item y = x\textsuperscript{2}/ 2p \\

 

Integrating from 0 to 2p 

\\\Rightarrow \int_{0}^{2 p} y d x=\int_{0}^{2 p} \frac{x^{2}}{2 p} d x \\ \Rightarrow \int_{0}^{2 p} y d x=\frac{1}{2 p}\left[\frac{x^{3}}{3}\right]_{0}^{2 p} \\ \Rightarrow \int_{0}^{2 p} y d x=\frac{1}{2 p}\left[\frac{(2 p)^{3}}{3}\right] \\ \Rightarrow \int_{0}^{2 p} y d x=\frac{(2 p)^{2}}{3} \\ \Rightarrow \int_{0}^{2 p} y d x=\frac{4 p^{2}}{3}

Using
(i)

\\\Rightarrow$ area bounded by two parabolas given $=\frac{8 \mathrm{p}^{2}}{3}-\frac{4 \mathrm{p}^{2}}{3}$ $\\\Rightarrow$ area bounded by two parabolas given $=\frac{4 \mathrm{p}^{2}}{3}$\\ Hence area is $\frac{4 \mathrm{p}^{2}}{3}$ unit ${ }^{2}$

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infoexpert22

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