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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

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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

\[{xy=c}^2 \]

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

\[x\frac{dy}{dx} + y = 0\]

\[ \Rightarrow \frac{dy}{dx} = \frac{- y}{x}\]

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

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

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

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

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

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

\[ \Rightarrow y t^2 - ct = - x + ct\]

\[ \Rightarrow x + y t^2 = 2ct\]

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

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

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

\[ \Rightarrow yt - c = t^3 x - c t^4 \]

\[ \Rightarrow x t^3 - yt = c t^4 - c\]

  Is there an error in this question or solution?
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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