Question

Show that the following set of curve intersect orthogonally y = x³ and 6y = 7 − x² ?

Answer 1

\[ y = x^3 . . . \left( 1 \right)\]

\[6y = 7 - x^2 . . . \left( 2 \right)\]

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

\[6 x^3 = 7 - x^2 \]

\[ \Rightarrow 6 x^3 + x^2 - 7 = 0\]

\[x=1 \text { satisfies this }.\]

\[\text { Dividing this byx-1 ,we get }\]

\[6 x^2 + 7x + 7 = 0, \]

\[ {\text { Discriminant } = 7}^2 -4\left( 6 \right)\left( 7 \right)=-119<0\]

\[\text { So, this has no real roots }.\]

\[\text { When} x=1,y= x^3 =1 (\text { From }(1))\]

\[\text { So,} \left( x, y \right)=\left( 1, 1 \right)\]

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

\[\frac{dy}{dx} = 3 x^2 \]

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

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

\[6\frac{dx}{dx} = - 2x\]

\[ \Rightarrow \frac{dx}{dx} = \frac{- 2x}{6} = \frac{- x}{3}\]

\[ \Rightarrow m_2 = \left( \frac{dy}{dx} \right)_\left( 1, 1 \right) = \frac{- 1}{3}\]

\[\text { Now,} m_1 \times m_2 = 3 \times \frac{- 1}{3}\]

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

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

So, the given curves intersect orthogonally.