Question

Show that the following set of curve intersect orthogonally x² + 4y² = 8 and x² − 2y² = 4 ?

Answer 1

\[ x^2 + 4 y^2 = 8 . . . \left( 1 \right)\]

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

\[\text { From (1) and (2) we get }\]

\[6 y^2 = 4\]

\[ \Rightarrow y^2 = \frac{2}{3}\]

\[ \Rightarrow y = \frac{\sqrt{2}}{\sqrt{3}} ory = \frac{- \sqrt{2}}{\sqrt{3}}\]

\[\text { From } (1),\]

\[ x^2 + \frac{8}{3} = 8\]

\[ \Rightarrow x^2 = \frac{16}{3}\]

\[ \Rightarrow x = \pm \frac{4}{\sqrt{3}}\]

\[\text { So },\left( x, y \right)=\left( \frac{4}{\sqrt{3}}, \frac{\sqrt{2}}{\sqrt{3}} \right),\left( \frac{4}{\sqrt{3}}, \frac{- \sqrt{2}}{\sqrt{3}} \right),\left( \frac{- 4}{\sqrt{3}}, \frac{\sqrt{2}}{\sqrt{3}} \right),\left( \frac{- 4}{\sqrt{3}}, - \frac{\sqrt{2}}{\sqrt{3}} \right)\]

\[\text { Consider point }\left( x_1 , y_1 \right)=\left( \frac{4}{\sqrt{3}}, \frac{\sqrt{2}}{\sqrt{3}} \right)\]

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

\[2x + 8y\frac{dy}{dx} = 0\]

\[ \Rightarrow \frac{dy}{dx} = \frac{- x}{4y}\]

\[ \Rightarrow m_1 = \left( \frac{dy}{dx} \right)_\left( \frac{4}{\sqrt{3}}, \frac{\sqrt{2}}{\sqrt{3}} \right) = \frac{- \frac{4}{\sqrt{3}}}{4\frac{\sqrt{2}}{\sqrt{3}}} = \frac{- 1}{\sqrt{2}}\]

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

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

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

\[ \Rightarrow m_2 = \left( \frac{dy}{dx} \right)_\left( \frac{4}{\sqrt{3}}, \frac{\sqrt{2}}{\sqrt{3}} \right) = \frac{\frac{4}{\sqrt{3}}}{2\frac{\sqrt{2}}{\sqrt{3}}} = \sqrt{2}\]

\[\text { Now,} m_1 \times m_2 = \frac{- 1}{\sqrt{2}} \times \sqrt{2}\]

\[ \Rightarrow m_1 \times m_2 = - 1\]

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

\[\text { Hence,, the curves are orthogonal at }\left( \frac{4}{\sqrt{3}}, \frac{\sqrt{2}}{\sqrt{3}} \right).\]

\[\text { Similarly, we can see that the curves are orthogonal in each possibility of }\left( x_1 , y_1 \right).\]