Advertisements
Advertisements
Question
Show that the curves 2x = y2 and 2xy = k cut at right angles, if k2 = 8 ?
Advertisements
Solution
\[\text { Given }: \]
\[2x = y^2 . . . \left( 1 \right)\]
\[2xy = k . . . \left( 2 \right)\]
\[\text { From (1) and (2), we get }\]
\[ y^3 = k\]
\[ \Rightarrow y = k^\frac{1}{3} \]
\[\text { From (1), we get }\]
\[2x = k^\frac{2}{3} \]
\[ \Rightarrow x = \frac{k^\frac{2}{3}}{2}\]
\[\text { On differentiating (1) w.r.t.x,we get }\]
\[2 = 2y\frac{dy}{dx}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{1}{y}\]
\[ \Rightarrow m_1 = \left( \frac{dy}{dx} \right)_\left( \frac{k^\frac{2}{3}}{2}, k^\frac{1}{3} \right) = \frac{1}{k^\frac{1}{3}} = k^\frac{- 1}{3} \]
\[\text { On differentiating (2) w.r.t.x,we get }\]
\[2x\frac{dy}{dx} + 2y = 0\]
\[ \Rightarrow \frac{dy}{dx} = \frac{- y}{x}\]
\[ \Rightarrow m_2 = \left( \frac{dy}{dx} \right)_\left( \frac{k^\frac{2}{3}}{2}, k^\frac{1}{3} \right) = \frac{- k^\frac{1}{3}}{\left( \frac{k^\frac{2}{3}}{2} \right)} = - 2 k^\frac{- 1}{3} \]
\[\text { It is given that the curves intersect orthogonally }.\]
\[ \therefore m_1 \times m_2 = - 1\]
\[ \Rightarrow k^\frac{- 1}{3} \times - 2 k^\frac{- 1}{3} = - 1\]
\[ \Rightarrow 2 k^\frac{- 2}{3} = 1\]
\[ \Rightarrow k^\frac{- 2}{3} = \frac{1}{2}\]
\[ \Rightarrow k^\frac{2}{3} = 2\]
\[\text { Cubing on both sides, we get }\]
\[ k^2 = 8\]
