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# Solution for Find the Equations of All Lines of Slope Zero and that Are Tangent to the Curve Y = 1 X 2 − 2 X + 3 ? - CBSE (Commerce) Class 12 - Mathematics

#### Question

Find the equations of all lines of slope zero and that are tangent to the curve $y = \frac{1}{x^2 - 2x + 3}$ ?

#### Solution

Slope of the given tangent is 0.

$\text { Let }\left( x_1 , y_1 \right)\text { be a point where the tangent is drawn to the curve} (1).$

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

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

$\text { Now,} y = \frac{1}{x^2 - 2x + 3}$

$\Rightarrow \frac{dy}{dx} = \frac{\left( x^2 - 2x + 3 \right)\left( 0 \right) - \left( 2x - 2 \right)1}{\left( x^2 - 2x + 3 \right)^2} = \frac{- 2x + 2}{\left( x^2 - 2x + 3 \right)^2}$

$\text { Slope of tangent }=\frac{- 2 x_1 + 2}{\left( {x_1}^2 - 2 x_1 + 3 \right)^2}$

$\text { Given that }$

$\text { Slope oftangent= slope of the given line }$

$\Rightarrow \frac{- 2 x_1 + 2}{\left( {x_1}^2 - 2 x_1 + 3 \right)^2} = 0$

$\Rightarrow - 2 x_1 + 2 = 0$

$\Rightarrow 2 x_1 = 2$

$\Rightarrow x_1 = 1$

$\text { Now }, y = \frac{1}{1 - 2 + 3} = \frac{1}{2} \left[ \text { From }\left( 1 \right) \right]$

$\therefore \left( x_1 , y_1 \right) = \left( 1, \frac{1}{2} \right)$

$\text { Equation oftangentis},$

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

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

$\Rightarrow y = \frac{1}{2}$

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#### Video TutorialsVIEW ALL [3]

Solution for question: Find the Equations of All Lines of Slope Zero and that Are Tangent to the Curve Y = 1 X 2 − 2 X + 3 ? concept: Tangents and Normals. For the courses CBSE (Commerce), CBSE (Arts), PUC Karnataka Science, CBSE (Science)
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