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Question
Show that the following curve intersect orthogonally at the indicated point x2 = y and x3 + 6y = 7 at (1, 1) ?
Solution
\[ x^2 = y . . . \left( 1 \right)\]
\[ x^3 + 6y = 7 . . . \left( 2 \right)\]
\[\text { Given point is }\left( 1, 1 \right)\]
\[\text { Differentiating (1) w.r.t.x, }\]
\[2x = \frac{dy}{dx}\]
\[ \Rightarrow m_1 = \left( \frac{dy}{dx} \right)_\left( 1, 1 \right) = 2\left( 1 \right) = 2\]
\[\text { Differentiating (2) w.r.t.x, }\]
\[3 x^2 + 6\frac{dy}{dx} = 0\]
\[ \Rightarrow \frac{dy}{dx} = \frac{- x^2}{2}\]
\[ \Rightarrow m_2 = \left( \frac{dy}{dx} \right)_\left( 1, 1 \right) = \frac{- 1}{2}\]
\[\text { Since,} m_1 \times m_2 = - 1\]
Hence, the given curves intersect orthogonally at the given point.
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