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Find the Point on the Curve Y = X2 − 2x + 3, Where the Tangent is Parallel to X-axis ? - Mathematics

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प्रश्न

Find the point on the curve y = x2 − 2x + 3, where the tangent is parallel to x-axis ?

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उत्तर

The slope of the x-axis is 0.
Now, let (x1y1) be the required point.
Here,

\[\text { Since, the point lies on the curve } . \]

\[\text { Hence,} y_1 = {x_1}^2 - 2 x_1 + 3 . . . \left( 1 \right)\]

\[\text { Now }, y = x^2 - 2x + 3\]

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

\[\text { Slope of the tangent at }\left( x, y \right)= \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) =2 x_1 - 2\]

\[\text { Given }:\]

\[\text { Slope of the tangent at }\left( x_1 , y_1 \right)=\text { Slope of thex-axis }\]

\[ = 2 x_1 - 2 = 0\]

\[ \Rightarrow x_1 = 1\]

\[\text { and }\]

\[ y_1 = 1 - 2 + 3 = 2 [\text { From } (1)]\]

\[ \therefore \text { Required point }=\left( x_1 , y_1 \right)=\left( 1, 2 \right)\]

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अध्याय 16: Tangents and Normals - Exercise 16.4 [पृष्ठ ४१]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 16 Tangents and Normals
Exercise 16.4 | Q 1 | पृष्ठ ४१

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