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Find the Equation of the Tangent and the Normal to the Following Curve at the Indicated Point Y = X4 − 6x3 + 13x2 − 10x + 5 at X = 1 ? - Mathematics

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प्रश्न

 Find the equation of the tangent and the normal to the following curve at the indicated point y = x4 − 6x3 + 13x2 − 10x + 5 at x = 1? 

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उत्तर

\[y= x^4 - 6 x^3 + 13 x^2 - 10x + 5\]

\[\text{ When }x = 1 , \]

`y = 1 - 6 + 13 - 10 + 5 = 3`

\[\text { So}, \left( x_1 , y_1 \right) = \left( 1, 3 \right)\]

\[\text { Now,} y= x^4 - 6 x^3 + 13 x^2 - 10x + 5\]

\[\text { Differentiating both sides w.r.t.x,} \]

\[\frac{dy}{dx} = 4 x^3 - 18 x^2 + 26x - 10\]

\[\text { Slope of tangent },m= \left( \frac{dy}{dx} \right)_\left( 1, 3 \right) =4-18+26 - 10 = 2\]

\[\text { Equation of tangent is },\]

\[y - y_1 = 2 \left( x - x_1 \right)\]

\[ \Rightarrow y - 3 = 2\left( x - 1 \right)\]

\[ \Rightarrow y - 3 = 2x - 2\]

\[ \Rightarrow 2x - y + 1 = 0\]

\[\text { Equation of normal is
},\]

\[y - y_1 = \frac{- 1}{m} \left( x - x_1 \right)\]

\[ \Rightarrow y - 3 = \frac{- 1}{2} \left( x - 1 \right)\]

\[ \Rightarrow 2y - 6 = - x + 1\]

\[ \Rightarrow x + 2y - 7 = 0\]

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पाठ 16: Tangents and Normals - Exercise 16.2 [पृष्ठ २७]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 16 Tangents and Normals
Exercise 16.2 | Q 3.02 | पृष्ठ २७

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