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If the Curves Y = 2 Ex and Y = Ae−X Intersect Orthogonally, Then a =

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प्रश्न

If the curves y = 2 ex and y = ae−x intersect orthogonally, then a = _____________ .

विकल्प

  • 1/2

  • −1/2

  • 2

  • 2e2

MCQ
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उत्तर

1/2

 

\[\text { Given }: \]

\[y = 2 e^x . . . \left( 1 \right)\]

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

\[\text { Let the point of intersection of these curves be }\left( x_1 , y_1 \right).\]

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

\[\frac{dy}{dx} = 2 e^x \]

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

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

\[\frac{dy}{dx} = - a e^{- x} \]

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

\[\text { It is given that the curves are orthogonal }.\]

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

\[ \Rightarrow 2 e^{x_1} \times \left( - a e^{- x_1} \right) = - 1\]

\[ \Rightarrow 2a = 1\]

\[ \Rightarrow a = \frac{1}{2}\]

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 15: Tangents and Normals - Exercise 16.5 [पृष्ठ ४३]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 15 Tangents and Normals
Exercise 16.5 | Q 20 | पृष्ठ ४३
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