Question

The curves y = ae-x and y = bex are orthogonal if
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Answer 1

SOLUTION :

TO DETERMINE

The condition for orthogonal for the curves

$\sf{ y = a {e}^{ - x} \: \: \: and \: \: \: y = b {e}^{x} }$

CONCEPT TO BE IMPLEMENTED

Two curves are said to be orthogonal if the tangents at the point of intersection is perpendicular

EVALUATION

Here the given equation of the curves are

$\sf{ y = a {e}^{ - x} \: \: \: .......(1) }$

$\sf{ y = b {e}^{x} } \: \: \: ........(2)$

$\sf{Let \: (x_1,y_1) \: be \: the \: point \: of \: intersection }$

Differentiating both sides of Equation (1) with respect to x we get

$\displaystyle \sf{ \frac{dy}{dx} = - a\: {e}^{ - x} }$

$\therefore \: \displaystyle \sf{ m_1 = \bigg [\frac{dy}{dx} \bigg ]_{(x_1,y_1)} = - a\: {e}^{ - x_1} }$

Differentiating both sides of Equation (2) with respect to x we get

$\displaystyle \sf{ \frac{dy}{dx} = b \: {e}^{ x} }$

$\therefore \: \displaystyle \sf{ m_2 = \bigg [\frac{dy}{dx} \bigg ]_{(x_2,y_2)} = b\: {e}^{ x_1} }$

Since the two curves are orthogonal

So

$\therefore \: \displaystyle \sf{ m_1 \times m_2\: = - 1 }$

$\implies \sf{ - ab \: {e}^{ (- x_1 + x_1) } = - 1}$

$\implies \sf{ ab = 1}$

Which is the required condition

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LEARN MORE FROM BRAINLY

The coordinates (x,y) of a particle moving along a

plane curve at any time t are given by

y'+ 2x = sin 2t , x' -2 y = cos 2t

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Answer 2

Step-by-step explanation:

1. the line y=x+1is trangent to the curve y2=4xat the point