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The Curves Y = Aex and Y = Be−X Cut Orthogonally, If - Mathematics

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Question

The curves y = aex and y = be−x cut orthogonally, if ___________ .

Options

  • a = b

  • a = −b

  • ab = 1

  • ab = 2

MCQ
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Solution

`ab = 1`

 

\[\text{ Given }: \]

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

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

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

\[\text { Now,} \]

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

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

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

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

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

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

\[\text { It is given that the curves cut orthogonally }.\]

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

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

\[ \Rightarrow ab = 1\]

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Chapter 16: Tangents and Normals - Exercise 16.5 [Page 43]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 16 Tangents and Normals
Exercise 16.5 | Q 18 | Page 43

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