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The angle of intersection between the curve \[{{x}^{2}}=8y\] and \[{{y}^{2}}=8x\] at \[\left( 0,0 \right)\] is:
(A) \[\dfrac{\pi }{4}\]
(B) \[\dfrac{\pi }{3}\]
(C) \[\dfrac{\pi }{6}\]
(D) \[\dfrac{\pi }{2}\]


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Answer
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Hint: First of all, differentiate the equations \[{{x}^{2}}=8y\] and \[{{y}^{2}}=8x\] with respect to x and the values of \[\dfrac{dy}{dx}\] after differentiating both of the equations. Now, put \[\left( 0,0 \right)\] in the values of \[\dfrac{dy}{dx}\] . We know that, \[\tan 0=0\] and \[\tan \dfrac{\pi }{2}=\infty \] . Now, solve it further and get the angle between the tangents.

Complete step-by-step answer:
According to the question, it is given that we have the equation of two curves,
\[{{x}^{2}}=8y\] …………………(1)
\[{{y}^{2}}=8x\] …………………..(2)
We have to find the angle of intersection between these two curves at the point whose coordinate is \[\left( 0,0 \right)\] .
For the angle of intersection between these two curves, we have to find the slope of the tangent at the point of intersection of these two curves.
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Let us find the slope of the tangent AB of the curve \[{{x}^{2}}=8y\] at the point \[\left( 0,0 \right)\] .
Differentiating with respect to x, the LHS and RHS of the equation of the curve \[{{x}^{2}}=8y\] , we get,
\[\dfrac{d{{x}^{2}}}{dx}=\dfrac{d\left( 8y \right)}{dx}\] ………………………(3)
We know the formula, \[\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}\] ……………………(4)
Using this formula and simplifying equation (3), we get
\[\dfrac{d{{x}^{2}}}{dx}=\dfrac{d\left( 8y \right)}{dx}\]
\[\begin{align}
  & \Rightarrow 2{{x}^{n-1}}=8\dfrac{dy}{dx} \\
 & \Rightarrow 2x=8\dfrac{dy}{dx} \\
 & \Rightarrow \dfrac{2x}{8}=\dfrac{dy}{dx} \\
\end{align}\]
\[\Rightarrow \dfrac{x}{4}=\dfrac{dy}{dx}\] …………………………(5)
We know that \[\dfrac{dy}{dx}\] is a slope which is a tan function. So,
\[\dfrac{dy}{dx}=\tan \theta =\dfrac{x}{4}\] ………………..(6)
We have to find the slope of the tangent AB at the point \[\left( 0,0 \right)\] . So, putting \[\left( 0,0 \right)\] in equation (6), we get,
\[\begin{align}
  & \Rightarrow \dfrac{x}{4}=\tan \theta \\
 & \Rightarrow \dfrac{0}{4}=\tan \theta \\
\end{align}\]
\[\Rightarrow 0=\tan \theta \] ………………..(7)
We know that, \[\tan 0=0\] ………………….(8)
Now, from equation (7) and equation (8), we get
\[\tan \theta =\tan 0{}^\circ \]
\[\Rightarrow \theta =0\] …………………..(9)
Now, let us find the slope of the tangent CD of the curve \[{{y}^{2}}=8x\] at the point \[\left( 0,0 \right)\] .
Differentiating with respect to x, the LHS and RHS of the equation of the curve \[{{y}^{2}}=8x\] , we get,
\[\dfrac{d{{y}^{2}}}{dx}=\dfrac{d\left( 8x \right)}{dx}\]
Using the chain rule, we can transform it as,
\[\Rightarrow \dfrac{d{{y}^{2}}}{dy}\times \dfrac{dy}{dx}=\dfrac{d\left( 8x \right)}{dx}\] ………………………(10)
We know the formula, \[\dfrac{d{{y}^{n}}}{dy}=n{{y}^{n-1}}\] ……………………(11)
Using this formula and simplifying equation (10), we get
\[\begin{align}
  & \Rightarrow \dfrac{d{{y}^{2}}}{dy}\times \dfrac{dy}{dx}=\dfrac{d\left( 8x \right)}{dx} \\
 & \Rightarrow 2{{y}^{2-1}}\times \dfrac{dy}{dx}=8 \\
 & \Rightarrow 2y\times \dfrac{dy}{dx}=8 \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{8}{2y} \\
\end{align}\]
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{4}{y}\] …………………………(12)
We know that \[\dfrac{dy}{dx}\] is a slope which is a tan function. So,
\[\dfrac{dy}{dx}=\tan \alpha =\dfrac{4}{y}\] ………………..(13)
We have to find the slope of the tangent CD at the point \[\left( 0,0 \right)\] . So, putting \[\left( 0,0 \right)\] in equation (13), we get,
\[\begin{align}
  & \Rightarrow \dfrac{4}{y}=\tan \alpha \\
 & \Rightarrow \dfrac{4}{0}=\tan \alpha \\
\end{align}\]
\[\Rightarrow \infty =\tan \alpha \] ………………..(14)
We know that, \[\tan \dfrac{\pi }{2}=\infty \] ………………….(15)
Now, from equation (14) and equation (15), we get
\[\tan \alpha =\tan \dfrac{\pi }{2}\]
\[\Rightarrow \alpha =\dfrac{\pi }{2}\] …………………..(16)
From equation (9) and equation (16), we have the angle of these tangents from the x-axis.
The angle between the tangents CD and AB = \[\alpha -\theta =\dfrac{\pi }{2}-0=\dfrac{\pi }{2}\] .
Therefore, the angle of intersection between the curves \[{{x}^{2}}=8y\] and \[{{y}^{2}}=8x\] at the point \[\left( 0,0 \right)\] is \[\dfrac{\pi }{2}\] .
Hence, the correct option is (D).

Note: In this question, one might think to apply the property that the product of the slope of two perpendicular lines is equal to -1.
The slope of the first tangent = \[\dfrac{x}{4}\] .
The slope of the second tangent = \[\dfrac{4}{y}\] .
The product of the slopes of these two tangents = \[\dfrac{x}{4}\times \dfrac{4}{y}=\dfrac{x}{y}\] …………….(1)
We have the coordinate of the point is \[\left( 0,0 \right)\] .
Putting \[\left( 0,0 \right)\] in equation (1), we get
The product of the slopes of these two tangents = \[\dfrac{x}{y}=\dfrac{0}{0}\] = undefined.
Therefore, here we cannot apply the property that the product of the slope of two perpendicular lines is equal to -1.