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Solution for The Equation of the Normal to the Curve Y = X(2 − X) at the Point (2, 0) is (A) X − 2y = 2 (B) X − 2y + 2 = 0 (C) 2x + Y = 4 (D) 2x + Y − 4 = 0 - CBSE (Commerce) Class 12 - Mathematics

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Question

The equation of the normal to the curve y = x(2 − x) at the point (2, 0) is
(a) x − 2y = 2
(b) x − 2y + 2 = 0
(c) 2x +  y = 4
(d) 2x + y − 4 = 0

Solution

(a) x − 2y = 2

\[\text { Here }, \]

\[y = x\left( 2 - x \right) = 2x - x^2 \]

\[ \Rightarrow \frac{dy}{dx} = 2 - 2x\]

\[\text { Slope of the tangent }= \left( \frac{dy}{dx} \right)_\left( 2, 0 \right) = 2 - 4 = - 2\]

\[\text { Slope of the normal }, m=\frac{- 1}{- 2}=\frac{1}{2}\]

\[\text { Given }: \]

\[\left( x_1 , y_1 \right) = \left( 2, 0 \right)\]

\[ \therefore \text { Equation of the normal }\]

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

\[ \Rightarrow y - 0 = \frac{1}{2}\left( x - 2 \right)\]

\[ \Rightarrow 2y = x - 2\]

\[ \Rightarrow x - 2y = 2\]

  Is there an error in this question or solution?
Solution The Equation of the Normal to the Curve Y = X(2 − X) at the Point (2, 0) is (A) X − 2y = 2 (B) X − 2y + 2 = 0 (C) 2x + Y = 4 (D) 2x + Y − 4 = 0 Concept: Tangents and Normals.
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