The orthogonal trajectory for the family of curves $e^{x} + e^{- y} = k$ is

e^{y} + e^{- x} = c

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e^{x} - e^{- y} = c

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e^{y} - e^{- x} = c

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None of the above

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Solution

The correct option is B $e^{y} - e^{- x} = c$
We have, $e^{x} + e^{- y} = k$ ....(1)
Differentiating w.r.t. $x$ , we get
$e^{x} - e^{- y} \frac{d y}{d x} = 0$ ....(2)
Replacing $\frac{d y}{d x}$ with $- \frac{d x}{d y}$ in (2), we get
$e^{x} + e^{- y} \frac{d x}{d y} = 0 ⟹ e^{y} d y = - e^{- x} d x$
On integrating, we get
$e^{y} = e^{- x} + c or e^{y} - e^{- x} = c$

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Similar questions

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The general solution of the differential equation $\frac{d y}{d x} = e^{x + y}$ is
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The orthogonal trajectory for the family of curves ex+e−y=k is

The correct option is B ey−e−x=cWe have, ex+e−y=k ....(1)Differentiating w.r.t. x, we getex−e−ydydx=0 ....(2)Replacing dydx with −dxdy in (2), we getex+e−ydxdy=0⟹eydy=−e−xdxOn integrating, we getey=e−x+c or ey−e−x=c

The orthogonal trajectory for the family of curves $e^{x} + e^{- y} = k$ is