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Find the value of $ {{a}^{2}} $ where two curves $ xy={{a}^{3}}\text{ and }x={{y}^{2}} $ cut each other orthogonally at a point.
(a)\[\dfrac{1}{3}\]
(b) \[3\]
(c) \[2\]
(d) \[\dfrac{1}{2}\]

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Last updated date: 13th Jun 2024
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Answer
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Hint: Here we will try to find the values of the slopes to the curves at any point. We know that the value of the slope of the tangent to a particular point on the curve is determined by the slope of the tangent at the point. The slope of the tangent to the curve is determined by differentiation. If two curves cut each other orthogonally then the tangents also intersect at right angles (orthogonally.).Then use the fact that the product of the slopes of two lines that intersect orthogonally is -1. \[\]

Complete step-by-step answer:
The given equations of the curves are
\begin{align*}
& x={{y}^{2}}.........\left( 1 \right) \\
& xy={{a}^{3}}........\left( 2 \right)
\end{align*}
Putting value of $ x $ form the equation (1) in equation (2),
\[\begin{align}
  & y\cdot {{y}^{2}}={{y}^{3}}={{a}^{3}} \\
 & \Rightarrow y=a \\
 & \therefore x={{y}^{2}}={{a}^{2}} \\
\end{align}\]
So the coordinates of the point of intersection are $ \left( {{a}^{2}},a \right) $ .
Now we will determine the value of slope of tangent to the curve at $ \left( {{a}^{2}},a \right) $ from equation (1),
\[\begin{align}
  & x={{y}^{2}} \\
 & \Rightarrow \dfrac{dx}{dx}=2y\dfrac{dy}{dx}\text{ }\left( \text{differentiating bothside wrt }x \right) \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{2y} \\
\end{align}\]
So the slope at $ \left( {{a}^{2}},a \right) $ is $ \dfrac{1}{2y}=\dfrac{1}{2a}={{m}_{1}}\left( \text{say} \right). $ \[\]
Similarly the value of the slope of the tangent to the curve at $ \left( {{a}^{2}},a \right) $ in equation (2),

\[\begin{align}
  & \text{ }xy={{a}^{3}} \\
 & \Rightarrow \dfrac{d}{dx}\left( xy \right)=0\left( \text{differentiating bothside wrt }x \right) \\
 & \Rightarrow x\dfrac{dy}{dx}+y\dfrac{dx}{dx}=0 \\
 & \Rightarrow \dfrac{dy}{dx}=-\dfrac{y}{x} \\
\end{align}\]
So the slope at $ \left( {{a}^{2}},a \right) $ is $ \dfrac{-y}{x}=\dfrac{-a}{{{a}^{2}}}=-\dfrac{1}{2}={{m}_{2}}\left( \text{say} \right) $ \[\]
If two curves cut orthogonally then the product of their slopes of the tangents at the point of intersection is -1.
\[\begin{align}
  & \therefore {{m}_{1}}{{m}_{2}}=-1 \\
 & \Rightarrow \dfrac{1}{2a}\cdot \left( \dfrac{-1}{a} \right)=-1 \\
 & \Rightarrow \dfrac{1}{2}={{a}^{2}} \\
\end{align}\]
So the only correct option is (D) $ \dfrac{1}{2} $ . We can verify this by plotting the curves taking $ {{a}^{2}}=\dfrac{1}{\sqrt{2}} $ and observe that the point of intersection is $ \left( {{a}^{2}},a \right)=\left( \dfrac{1}{2},\dfrac{1}{\sqrt{2}} \right) $ .
So, the correct answer is “Option D”.

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Note: The question tests your knowledge of implicit differentiation and tangent of the curves. Be careful when you differentiate $ x={{y}^{2}} $ , because if you go the route of direct differentiation and take square root on both sides that will give you two results.