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The Equation to the Normal to the Curve Y = Sin X at (0, 0) is - Mathematics

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प्रश्न

The equation to the normal to the curve y = sin x at (0, 0) is ___________ .

पर्याय

  • x = 0

  • y = 0

  • x + y = 0

  • x − y = 0

MCQ
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उत्तर

x + y = 0

\[\text { Given }: \]

\[y = \sin x\]

\[\text { On differentiating both sides w.r.t.x, we get }\]

\[\frac{dy}{dx} = \cos x\]

\[\text { Slope of the tangent }= \left( \frac{dy}{dx} \right)_\left( 0, 0 \right) = cos 0 =1\]

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

\[\text { Given }: \]

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

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

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

\[ \Rightarrow y - 0 = - 1\left( x - 0 \right)\]

\[ \Rightarrow y = - x\]

\[ \Rightarrow x + y = 0\]

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पाठ 16: Tangents and Normals - Exercise 16.5 [पृष्ठ ४२]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 16 Tangents and Normals
Exercise 16.5 | Q 1 | पृष्ठ ४२

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