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प्रश्न
Find the equation of the normal to the curve x2 + 2y2 − 4x − 6y + 8 = 0 at the point whose abscissa is 2 ?
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उत्तर
Abscissa means the horizontal co-ordiante of a point.
Given that abscissa = 2.
i.e., x = 2
\[x^2 + 2 y^2 - 4x - 6y + 8 = 0 . . . \left( 1 \right)\]
\[\text { Differentiating both sides w.r.t.x }, \]
\[2x + 4y\frac{dy}{dx} - 4 - 6\frac{dy}{dx} = 0\]
\[ \Rightarrow \frac{dy}{dx}\left( 4y - 6 \right) = 4 - 2x\]
\[ \Rightarrow \frac{dy}{dx} = \frac{4 - 2x}{4y - 6} = \frac{2 - x}{2y - 3}\]
\[\text { When }x=2,\text { from } (1), \text { we get }\]
\[4 + 2 y^2 - 8 - 6y + 8 = 0\]
\[ \Rightarrow 2 y^2 - 6y + 4 = 0\]
\[ \Rightarrow y^2 - 3y + 2 = 0\]
\[ \Rightarrow \left( y - 1 \right)\left( y - 2 \right) = 0\]
\[ \Rightarrow y = 1 ory = 2\]
\[\text { Case }-1:y = 1\]
\[\text { Slope of tangent } = \left( \frac{dy}{dx} \right)_\left( 2, 1 \right) =\frac{0}{- 1}=0\]
\[\left( x_1 , y_1 \right) = \left( 2, 1 \right)\]
\[\text { Equation of normal is },\]
\[y - y_1 = \frac{- 1}{m} \left( x - x_1 \right)\]
\[ \Rightarrow y - 1 = \frac{- 1}{0} \left( x - 2 \right)\]
\[ \Rightarrow x - 2 = 0\]
\[ \Rightarrow x = 2\]
\[\text { Case}-2:y = 2\]
\[\text { Slope of tangent} = \left( \frac{dy}{dx} \right)_\left( 2, 2 \right) =\frac{0}{1}=0\]
\[\left( x_1 , y_1 \right) = \left( 2, 2 \right)\]
\[\text { Equation of normal is },\]
\[y - y_1 = \frac{- 1}{m} \left( x - x_1 \right)\]
\[ \Rightarrow y - 2 = \frac{- 1}{0} \left( x - 2 \right)\]
\[ \Rightarrow x - 2 = 0\]
\[ \Rightarrow x = 2\]
In both cases, the equation of normal is x = 2
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