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Find the Equation of the Tangent and the Normal to the Following Curve at the Indicated Point Y2 = 4ax at (X1, Y1)? - Mathematics

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प्रश्न

Find the equation of the tangent and the normal to the following curve at the indicated point  y2 = 4ax at (x1, y1)?

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उत्तर

\[y^2 =4ax\]

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

\[2y\frac{dy}{dx} = 4a\]

\[ \Rightarrow \frac{dy}{dx} = \frac{2a}{y}\]

\[\text { At }\left( x_1 , y_1 \right)\]

\[\text { Slope of tangent } = \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) =\frac{2a}{y_1}=m\]

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

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

\[ \Rightarrow y - y_1 = \frac{2a\left( x - x_1 \right)}{y_1}\]

\[ \Rightarrow y y_1 - {y_1}^2 = 2ax - 2a x_1 \]

\[ \Rightarrow y y_1 - 4a x_1 = 2ax - 2a x_1 \]

\[ \Rightarrow y y_1 = 2ax + 2a x_1 \]

\[ \Rightarrow y y_1 = 2a\left( x + x_1 \right)\]

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

\[y - y_1 = \frac{1}{\text { Slope of tangen}t} \left( x - x_1 \right)\]

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

\[ \Rightarrow y - y_1 = \frac{- y_1}{2a}\left( x - x_1 \right)\]

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

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 16 Tangents and Normals
Exercise 16.2 | Q 3.18 | पृष्ठ २७

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