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# Solution for Find the Equation of the Tangent and the Normal to the Following Curve at the Indicated Point Xy = C2 at ( C T , C T ) ? - CBSE (Commerce) Class 12 - Mathematics

#### Question

Find the equation of the tangent and the normal to the following curve at the indicated point xy = c2 at $\left( ct, \frac{c}{t} \right)$ ?

#### Solution

${xy=c}^2$

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

$x\frac{dy}{dx} + y = 0$

$\Rightarrow \frac{dy}{dx} = \frac{- y}{x}$

$\text { Given } \left( x_1 , y_1 \right) = \left( ct, \frac{c}{t} \right)$

$\text { Slope of tangent,}m= \left( \frac{dy}{dx} \right)_\left( ct, \frac{c}{t} \right) =\frac{- \frac{c}{t}}{ct}=\frac{- 1}{t^2}$

$\text { Equation of tangent is },$

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

$\Rightarrow y - \frac{c}{t} = \frac{- 1}{t^2} \left( x - ct \right)$

$\Rightarrow \frac{yt - c}{t} = \frac{- 1}{t^2} \left( x - ct \right)$

$\Rightarrow y t^2 - ct = - x + ct$

$\Rightarrow x + y t^2 = 2ct$

$\text{ Equation of normal is },$

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

$\Rightarrow y - \frac{c}{t} = t^2 \left( x - ct \right)$

$\Rightarrow yt - c = t^3 x - c t^4$

$\Rightarrow x t^3 - yt = c t^4 - c$

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#### Video TutorialsVIEW ALL [3]

Solution for question: Find the Equation of the Tangent and the Normal to the Following Curve at the Indicated Point Xy = C2 at ( C T , C T ) ? concept: Tangents and Normals. For the courses CBSE (Commerce), CBSE (Arts), PUC Karnataka Science, CBSE (Science)
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