Answer
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Hint: In this problem, first we need to draw the curve of two parabolas. Now, find the first derivative of the given parabolas. Next, find the angle between the two parabolas at the origin.
Complete step-by-step answer:
The graph of the parabolas \[{y^2} = 4ax\] and \[{x^2} = 4ay\] is shown below.
Now, find the first derivative of the parabola \[{y^2} = 4ax\] with respect to \[x\] as shown below.
\[\begin{gathered}
\,\,\,\,\,\,\,\dfrac{d}{{dx}}\left( {{y^2}} \right) = \dfrac{d}{{dx}}\left( {4ax} \right) \\
\Rightarrow 2y\dfrac{{dy}}{{dx}} = 4a\dfrac{d}{{dx}}\left( x \right) \\
\Rightarrow 2y\dfrac{{dy}}{{dx}} = 4a \\
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{4a}}{{2y}} \\
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{2a}}{y} \\
\end{gathered}\]
From the above figure, it can be observed that, \[\dfrac{{dy}}{{dx}} \to \infty\], therefore,
\[\begin{gathered}
\,\,\,\,\,\tan {\theta _1} = \tan \dfrac{\pi }{2} \\
\Rightarrow {\theta _1} = \dfrac{\pi }{2} \\
\end{gathered}\]
Now, find the first derivative of the parabola \[{x^2} = 4ay\] with respect to \[x\] as shown below.
\[\begin{gathered}
\,\,\,\,\,\,\,\dfrac{d}{{dx}}\left( {{x^2}} \right) = \dfrac{d}{{dx}}\left( {4ay} \right) \\
\Rightarrow 2x = 4a\dfrac{{dy}}{{dx}} \\
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{2x}}{{4a}} \\
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{x}{{2a}} \\
\end{gathered}\]
From the above figure, it can be observed that, \[\dfrac{{dy}}{{dx}} \to 0\], therefore,
\[\begin{gathered}
\,\,\,\,\,\tan {\theta _2} = \tan 0 \\
\Rightarrow {\theta _2} = 0 \\
\end{gathered}\]
Now, the angle between the two curves is calculated as follows:
\[\begin{gathered}
\,\,\,\,\,\,\,{\theta _1} - {\theta _2} \\
\Rightarrow \dfrac{\pi }{2} - 0 \\
\Rightarrow \dfrac{\pi }{2} \\
\end{gathered}\]
Thus, option (C) is the correct answer.
Note: The parabola, \[{y^2} = 4ax\] makes an angle of 90 degree at the origin, whereas, the parabola \[{x^2} = 4ay\] makes an angle of 0 degree at origin.
Complete step-by-step answer:
The graph of the parabolas \[{y^2} = 4ax\] and \[{x^2} = 4ay\] is shown below.
![seo images](https://www.vedantu.com/question-sets/6e5aea3c-c86b-4642-8e56-fd74d8ae34163107958193463241785.png)
Now, find the first derivative of the parabola \[{y^2} = 4ax\] with respect to \[x\] as shown below.
\[\begin{gathered}
\,\,\,\,\,\,\,\dfrac{d}{{dx}}\left( {{y^2}} \right) = \dfrac{d}{{dx}}\left( {4ax} \right) \\
\Rightarrow 2y\dfrac{{dy}}{{dx}} = 4a\dfrac{d}{{dx}}\left( x \right) \\
\Rightarrow 2y\dfrac{{dy}}{{dx}} = 4a \\
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{4a}}{{2y}} \\
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{2a}}{y} \\
\end{gathered}\]
From the above figure, it can be observed that, \[\dfrac{{dy}}{{dx}} \to \infty\], therefore,
\[\begin{gathered}
\,\,\,\,\,\tan {\theta _1} = \tan \dfrac{\pi }{2} \\
\Rightarrow {\theta _1} = \dfrac{\pi }{2} \\
\end{gathered}\]
Now, find the first derivative of the parabola \[{x^2} = 4ay\] with respect to \[x\] as shown below.
\[\begin{gathered}
\,\,\,\,\,\,\,\dfrac{d}{{dx}}\left( {{x^2}} \right) = \dfrac{d}{{dx}}\left( {4ay} \right) \\
\Rightarrow 2x = 4a\dfrac{{dy}}{{dx}} \\
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{2x}}{{4a}} \\
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{x}{{2a}} \\
\end{gathered}\]
From the above figure, it can be observed that, \[\dfrac{{dy}}{{dx}} \to 0\], therefore,
\[\begin{gathered}
\,\,\,\,\,\tan {\theta _2} = \tan 0 \\
\Rightarrow {\theta _2} = 0 \\
\end{gathered}\]
Now, the angle between the two curves is calculated as follows:
\[\begin{gathered}
\,\,\,\,\,\,\,{\theta _1} - {\theta _2} \\
\Rightarrow \dfrac{\pi }{2} - 0 \\
\Rightarrow \dfrac{\pi }{2} \\
\end{gathered}\]
Thus, option (C) is the correct answer.
Note: The parabola, \[{y^2} = 4ax\] makes an angle of 90 degree at the origin, whereas, the parabola \[{x^2} = 4ay\] makes an angle of 0 degree at origin.
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