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The orthogonal trajectories of the family of curves y = C1x3 are
  • a)
    2 x3 + 3 y2 = C2
  • b)
    3 x2 + y2 = C2
  • c)
    3 x2 + 2 y2 = C2
  • d)
    x2 + 3 y2 = C2
Correct answer is option 'D'. Can you explain this answer?
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The orthogonal trajectories of the family of curves y = C1x3 area)2 x3...
Orthogonal Trajectories of a Family of Curves

Orthogonal trajectories are the curves that intersect another family of curves at right angles. These curves are useful in various fields like physics, engineering, and mathematics. To find the orthogonal trajectories of a family of curves, we follow the below steps:

Steps

1. Find the derivative of the given family of curves.

2. Take the negative reciprocal of the derivative to get the slope of the orthogonal curve.

3. Use this slope to find the equation of the orthogonal curve.

Orthogonal Trajectories of Given Families of Curves

Let's find the orthogonal trajectories of the given families of curves:

a) y = C1x^3 - 2x^3

1. Differentiating with respect to x, we get:

dy/dx = 3C1x^2 - 6x^2

2. Taking the negative reciprocal, we get:

slope of orthogonal curve = -1/(dy/dx) = (3C1x^2 - 6x^2)^-1

3. Using this slope, we can find the equation of the orthogonal curve. However, since this equation is very complex, we can't simplify it further.

b) 3x^2 - y^2 = C2

1. Differentiating with respect to x, we get:

dy/dx = -6x/y

2. Taking the negative reciprocal, we get:

slope of orthogonal curve = -1/(dy/dx) = y/(6x)

3. Substituting this slope in the equation of the given family of curves, we get:

3x^2 - y^2 = C2

Now, taking the derivative of this equation with respect to x, we get:

6x - 2yy' = 0

Solving this equation for y', we get:

y' = 3x/y

This is the equation of the orthogonal curve.

c) 3x^2 - 2y^2 = C2

1. Differentiating with respect to x, we get:

dy/dx = -3x/2y

2. Taking the negative reciprocal, we get:

slope of orthogonal curve = -1/(dy/dx) = 2y/(3x)

3. Substituting this slope in the equation of the given family of curves, we get:

3x^2 - 2y^2 = C2

Now, taking the derivative of this equation with respect to x, we get:

6x - 4yy' = 0

Solving this equation for y', we get:

y' = 3x/(2y)

This is the equation of the orthogonal curve.

d) x^2 - 3y^2 = C2

1. Differentiating with respect to x, we get:

dy/dx = -x/(3y)

2. Taking the negative reciprocal, we get:

slope of orthogonal curve = -1/(dy/dx) = 3y/x

3. Substituting this slope in the equation of the given family of curves, we get:

x^2 - 3y^2 = C2

Now, taking the derivative of this equation with respect to x, we get:

2x - 6yy' = 0

Solving this equation for y', we get:

y' = x/(3y)

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The orthogonal trajectories of the family of curves y = C1x3 area)2 x3 + 3 y2 = C2b)3 x2 + y2 = C2c)3 x2 + 2 y2 = C2d)x2 + 3 y2 = C2Correct answer is option 'D'. Can you explain this answer?
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The orthogonal trajectories of the family of curves y = C1x3 area)2 x3 + 3 y2 = C2b)3 x2 + y2 = C2c)3 x2 + 2 y2 = C2d)x2 + 3 y2 = C2Correct answer is option 'D'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about The orthogonal trajectories of the family of curves y = C1x3 area)2 x3 + 3 y2 = C2b)3 x2 + y2 = C2c)3 x2 + 2 y2 = C2d)x2 + 3 y2 = C2Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The orthogonal trajectories of the family of curves y = C1x3 area)2 x3 + 3 y2 = C2b)3 x2 + y2 = C2c)3 x2 + 2 y2 = C2d)x2 + 3 y2 = C2Correct answer is option 'D'. Can you explain this answer?.
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