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Question
Show that the following curve intersect orthogonally at the indicated point x2 = 4y and 4y + x2 = 8 at (2, 1) ?
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Solution
\[ x^2 = 4y . . . \left( 1 \right)\]
\[4y + x^2 = 8 . . . \left( 2 \right)\]
\[\text { Given point is }\left( 2, 1 \right)\]
\[\text { Differentiating (1) w.r.t.x,}\]
\[2x = 4\frac{dy}{dx}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{x}{2}\]
\[ \Rightarrow m_1 = \left( \frac{dy}{dx} \right)_\left( 2, 1 \right) = \frac{2}{2} = 1\]
\[\text { Differentiating (2) w.r.t.x,}\]
\[4\frac{dy}{dx} + 2x = 0\]
\[ \Rightarrow \frac{dy}{dx} = \frac{- x}{2}\]
\[ \Rightarrow m_2 = \left( \frac{dy}{dx} \right)_\left( 2, 1 \right) = \frac{- 2}{2} = - 1\]
\[\text { Since, } m_1 \times m_2 = - 1\]
Hence, the given curves intersect orthogonally at the given point.
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