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Question
The equation of the normal to the curve y = x(2 − x) at the point (2, 0) is ________________ .
Options
x − 2y = 2
x − 2y + 2 = 0
2x + y = 4
2x + y − 4 = 0
Solution
x − 2y = 2
\[\text { Here }, \]
\[y = x\left( 2 - x \right) = 2x - x^2 \]
\[ \Rightarrow \frac{dy}{dx} = 2 - 2x\]
\[\text { Slope of the tangent }= \left( \frac{dy}{dx} \right)_\left( 2, 0 \right) = 2 - 4 = - 2\]
\[\text { Slope of the normal }, m=\frac{- 1}{- 2}=\frac{1}{2}\]
\[\text { Given }: \]
\[\left( x_1 , y_1 \right) = \left( 2, 0 \right)\]
\[ \therefore \text { Equation of the normal }\]
\[ = y - y_1 = m\left( x - x_1 \right)\]
\[ \Rightarrow y - 0 = \frac{1}{2}\left( x - 2 \right)\]
\[ \Rightarrow 2y = x - 2\]
\[ \Rightarrow x - 2y = 2\]
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