Question

Find the equation of the tangent and the normal to the following curve at the indicated point \[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \text { at } \left( x_0 , y_0 \right)\] ?

Answer 1

\[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\]

\[\text { Differentiating both sides w.r.t.x,} \]

\[\frac{2x}{a^2} - \frac{2y}{b^2}\frac{dy}{dx} = 0\]

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

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

\[\text { Slope of tangent,}m= \left( \frac{dy}{dx} \right)_\left( x_0 , y_0 \right) =\frac{x_0 b^2}{y_0 a^2}\]

\[\text { Equation of tangent is },\]

\[y - y_1 = m \left( x - x_1 \right)\]

\[ \Rightarrow y - y_0 = \frac{x_0 b^2}{y_0 a^2}\left( x - x_0 \right)\]

\[ \Rightarrow y y_0 a^2 - {y_0}^2 a^2 = x x_0 b^2 - {x_0}^2 b^2 \]

\[x x_0 b^2 - y y_0 a^2 = {x_0}^2 b^2 - {y_0}^2 a^2 . . . \left( 1 \right)\]

\[\text { Since }\left( x_0 , y_0 \right)\text { lies on the given curve},\]

\[ \Rightarrow \frac{{x_0}^2}{a^2} - \frac{{y_0}^2}{b^2} = 1\]

\[ \Rightarrow {x_0}^2 b^2 - {y_0}^2 a^2 = a^2 b^2 \]

\[\text { Substituting this in (1), we get }\]

\[ \Rightarrow x x_0 b^2 - y y_0 a^2 = a^2 b^2 \]

\[\text { Dividing this by} a^2 b^2 \]

\[\frac{x x_0}{a^2} - \frac{y y_0}{b^2} = 1\]

\[\text { Equation of normal is,}\]

\[y - y_1 = m \left( x - x_1 \right)\]

\[ \Rightarrow y - y_0 = \frac{- y_0 a^2}{x_0 b^2}\left( x - x_0 \right)\]

\[ \Rightarrow y x_0 b^2 - x_0 y_0 b^2 = - x y_0 a^2 + x_0 y_0 a^2 \]

\[ \Rightarrow x y_0 a^2 + y x_0 b^2 = x_0 y_0 a^2 + x_0 y_0 b^2 \]

\[ \Rightarrow x y_0 a^2 + y x_0 b^2 = x_0 y_0 \left( a^2 + b^2 \right)\]

\[\text { Dividing by } x_0 y_0 \]

\[\frac{a^2 x}{x_0} + \frac{b^2 y}{y_0} = a^2 + b^2\]