Question

Find the slope of the tangent and the normal to the following curve at the indicted point  x² + 3y + y² = 5 at (1, 1)  ?

Answer 1

\[x^2 + 3y + y^2 = 5\]

\[\text { On differentiating both sides w.r.t.x, we get }\]

\[2x + 3\frac{dy}{dx} + 2y \frac{dy}{dx} = 0\]

\[ \Rightarrow \frac{dy}{dx}\left( 3 + 2y \right) = - 2x\]

\[ \Rightarrow \frac{dy}{dx} = \frac{- 2x}{3 + 2y}\]

\[\text { Now,} \]

\[\text { Slope of the tangent }= \left( \frac{dy}{dx} \right)_\left( 1, 1 \right) =\frac{- 2x}{3 + 2y}=\frac{- 2}{3 + 2}=\frac{- 2}{5}\]

\[\text { Slope of the normal }=\frac{- 1}{\left( \frac{dy}{dx} \right)_\left( 1, 1 \right)}=\frac{- 1}{\left( \frac{- 2}{5} \right)}=\frac{5}{2}\]