Question

Show that the following curve intersect orthogonally at the indicated point x² = 4y and 4y + x² = 8 at (2, 1) ?

Answer 1

\[ x^2 = 4y . . . \left( 1 \right)\]

\[4y + x^2 = 8 . . . \left( 2 \right)\]

\[\text { Given point is }\left( 2, 1 \right)\]

\[\text { Differentiating (1) w.r.t.x,}\]

\[2x = 4\frac{dy}{dx}\]

\[ \Rightarrow \frac{dy}{dx} = \frac{x}{2}\]

\[ \Rightarrow m_1 = \left( \frac{dy}{dx} \right)_\left( 2, 1 \right) = \frac{2}{2} = 1\]

\[\text { Differentiating (2) w.r.t.x,}\]

\[4\frac{dy}{dx} + 2x = 0\]

\[ \Rightarrow \frac{dy}{dx} = \frac{- x}{2}\]

\[ \Rightarrow m_2 = \left( \frac{dy}{dx} \right)_\left( 2, 1 \right) = \frac{- 2}{2} = - 1\]

\[\text { Since, } m_1 \times m_2 = - 1\]

Hence,  the given curves intersect orthogonally at the given point.