मराठी

Find the Condition for the Following Set of Curve to Intersect Orthogonally X 2 a 2 − Y 2 B 2 = 1 and X Y = C 2 ? - Mathematics

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प्रश्न

Find the condition for the following set of curve to intersect orthogonally \[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \text { and } xy = c^2\] ?

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उत्तर

\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 . . . \left( 1 \right)\]

\[xy = c^2 . . . \left( 2 \right)\]

\[\text { Let the curves intersect orthogonally at }\left( x_1 , y_1 \right).\]

\[\text { On differentiating (1) on both sides w.r.t.x, we get }\]

\[\frac{2x}{a^2} - \frac{2y}{b^2}\frac{dy}{dx} = 0\]

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

\[ \Rightarrow m_1 = \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) = \frac{x_1 b^2}{a^2 y_1}\]

\[\text { On differentiating (2) on both sides w.r.t.x, we get }\]

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

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

\[ \Rightarrow m_2 = \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) = \frac{- y_1}{x_1}\]

\[\text { It is given that the curves intersect orhtogonally at }\left( x_1 , y_1 \right).\]

\[ \therefore m_1 \times m_2 = - 1\]

\[ \Rightarrow \frac{x_1 b^2}{a^2 y_1} \times \frac{- y_1}{x_1} = - 1\]

\[ \Rightarrow a^2 = b^2 \]

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पाठ 16: Tangents and Normals - Exercise 16.3 [पृष्ठ ४१]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 16 Tangents and Normals
Exercise 16.3 | Q 8.1 | पृष्ठ ४१

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