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Question
Show that the following set of curve intersect orthogonally y = x3 and 6y = 7 − x2 ?
Solution
\[ y = x^3 . . . \left( 1 \right)\]
\[6y = 7 - x^2 . . . \left( 2 \right)\]
\[\text { From (1) and (2) we get }\]
\[6 x^3 = 7 - x^2 \]
\[ \Rightarrow 6 x^3 + x^2 - 7 = 0\]
\[x=1 \text { satisfies this }.\]
\[\text { Dividing this byx-1 ,we get }\]
\[6 x^2 + 7x + 7 = 0, \]
\[ {\text { Discriminant } = 7}^2 -4\left( 6 \right)\left( 7 \right)=-119<0\]
\[\text { So, this has no real roots }.\]
\[\text { When} x=1,y= x^3 =1 (\text { From }(1))\]
\[\text { So,} \left( x, y \right)=\left( 1, 1 \right)\]
\[\text { Differentiating (1) w.r.t.x, }\]
\[\frac{dy}{dx} = 3 x^2 \]
\[ \Rightarrow m_1 = \left( \frac{dy}{dx} \right)_\left( 1, 1 \right) = 3\]
\[\text { Differenntiating (2) w.r.t.x, }\]
\[6\frac{dx}{dx} = - 2x\]
\[ \Rightarrow \frac{dx}{dx} = \frac{- 2x}{6} = \frac{- x}{3}\]
\[ \Rightarrow m_2 = \left( \frac{dy}{dx} \right)_\left( 1, 1 \right) = \frac{- 1}{3}\]
\[\text { Now,} m_1 \times m_2 = 3 \times \frac{- 1}{3}\]
\[ \Rightarrow m_1 \times m_2 = - 1\]
\[\text { Since }, m_1 \times m_2 = - 1\]
So, the given curves intersect orthogonally.
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