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प्रश्न
Find the equation of the tangent and the normal to the following curve at the indicated point \[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \text { at } \left( \sqrt{2}a, b \right)\] ?
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उत्तर
\[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\]
\[\text { Differentiating both sides w.r.t.x,} \]
\[\frac{2x}{a^2} - \frac{2y}{b^2}\frac{dy}{dx} = 0\]
\[ \Rightarrow \frac{2y}{b^2}\frac{dy}{dx} = \frac{2x}{a^2}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{x b^2}{y a^2}\]
\[\text { Slope of tangent },m= \left( \frac{dy}{dx} \right)_\left( \sqrt{2}a,b \right) =\frac{\sqrt{2}a b^2}{b a^2}=\frac{\sqrt{2}b}{a}\]
\[\text { Equation of tangent is },\]
\[y - y_1 = m\left( x - x_1 \right)\]
\[ \Rightarrow y - b = \frac{\sqrt{2}b}{a}\left( x - \sqrt{2}a \right)\]
\[ \Rightarrow ay - ab = \sqrt{2}bx - 2ab\]
\[ \Rightarrow \sqrt{2}bx - ay = ab\]
\[ \Rightarrow \frac{\sqrt{2}x}{a} - \frac{y}{b} = 1\]
\[\text { Equation of normal is, }\]
\[y - y_1 = \frac{- 1}{m}\left( x - x_1 \right)\]
\[ \Rightarrow y - b = \frac{- a}{\sqrt{2}b}\left( x - \sqrt{2}a \right)\]
\[ \Rightarrow \sqrt{2}by - \sqrt{2} b^2 = - ax + \sqrt{2} a^2 \]
\[ \Rightarrow ax + \sqrt{2}by = \sqrt{2} b^2 + \sqrt{2} a^2 \]
\[ \Rightarrow \frac{ax}{\sqrt{2}} + by = a^2 + b^2\]
