Question

Find the equation of the tangent and the normal to the following curve at the indicated point \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \text { at } \left( x_1 , y_1 \right)\] ?

Answer 1

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

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

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

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

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

\[\text { Slope of tangent,}m= \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) =\frac{- x_1 b^2}{y_1 a^2}\]

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

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

\[ \Rightarrow y - y_1 = \frac{- x_1 b^2}{y_1 a^2}\left( x - x_1 \right)\]

\[ \Rightarrow y y_1 a^2 - {y_1}^2 a^2 = - x x_1 b^2 + {x_1}^2 b^2 \]

\[ \Rightarrow x x_1 b^2 + y y_1 a^2 = {x_1}^2 b^2 + {y_1}^2 a^2 . . . \left( 1 \right)\]

\[\text { Since }\left( x_1 , y_1 \right)\text { lies on the given curve.Therefore},\]

\[\frac{{x_1}^2}{a^2} + \frac{{y_1}^2}{b^2} = 1\]

\[ \Rightarrow \frac{{x_1}^2 b^2 + {y_1}^2 a^2}{a^2 b^2} = 1\]

\[ \Rightarrow {x_1}^2 b^2 + {y_1}^2 a^2 = a^2 b^2 \]

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

\[x x_1 b^2 + y y_1 a^2 = a^2 b^2 \]

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

\[\frac{x x_1}{a^2} + \frac{y y_1}{b^2} = 1\]

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

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

\[ \Rightarrow y - y_1 = \frac{y_1 a^2}{x_1 b^2}\left( x - x_1 \right)\]

\[ \Rightarrow y x_1 b^2 - x_1 y_1 b^2 = x y_1 a^2 - x_1 y_1 a^2 \]

\[ \Rightarrow x y_1 a^2 - y x_1 b^2 = x_1 y_1 a^2 - x_1 y_1 b^2 \]

\[ \Rightarrow x y_1 a^2 - y x_1 b^2 = x_1 y_1 \left( a^2 - b^2 \right)\]

\[\text { Dividing by } x_1 y_1 \]

\[\frac{a^2 x}{x_1} - \frac{b^2 y}{y_1} = a^2 - b^2\]