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Answer :
The orthogonal trajectories are given by options (C), (F), and (G), i.e.,
[tex]\(y^2 + 2x^2 = C\),[/tex]
[tex]\(2y^3 + 25x^3 = C\)[/tex], and
[tex]\(23y^2 + 23x^2 = C\)[/tex].
To find the orthogonal trajectories of the family of curves given by, we need to find the differential equation satisfied by the orthogonal trajectories and then solve it to obtain the desired equations.
Let's start by finding the differential equation for the family of curves [tex]\(y^4 = kx^3\)[/tex]. Differentiating both sides with respect to (x) gives:
[tex]\[4y^3 \frac{dy}{dx} = 3kx^2.\][/tex]
Now, we can find the slope of the tangent line for the family of curves. The slope of the tangent line is given by [tex]\(\frac{dy}{dx}\)[/tex], and the slope of the orthogonal trajectory will be the negative reciprocal of this slope.
So, the slope of the orthogonal trajectory is
[tex]\(-\frac{1}{4y^3} \cdot \frac{dx}{dy}\).[/tex]
To find the differential equation satisfied by the orthogonal trajectories, we equate the negative reciprocal of the slope to the derivative of \(y\) with respect to \(x\):
[tex]\[-\frac{1}{4y^3} \cdot \frac{dx}{dy} = \frac{dy}{dx}.\][/tex]
Simplifying this equation, we get:
[tex]\[-\frac{1}{4y^3} dy = dx.\][/tex]
Now, we integrate both sides with respect to the respective variables:
[tex]\[-\int \frac{1}{4y^3} dy = \int dx.\][/tex]
Integrating, we have:
[tex]\[\frac{1}{12y^2} = x + C,\][/tex]
where (C) is the constant of integration.
This equation represents the orthogonal trajectories of the family of curves [tex]\(y^4 = kx^3\)[/tex].
Let's check which of the given options satisfy the equation
[tex]\(\frac{1}{12y^2} = x + C\):[/tex]
(A) [tex]\(25y^3 + 3x^2 = C\)[/tex] does not satisfy the equation.
(B) [tex]\(2y^3 + 2x^2 = C\)[/tex] does not satisfy the equation.
(C) [tex]\(y^2 + 2x^2 = C\)[/tex] satisfies the equation with [tex]\(C = \frac{1}{12}\)[/tex].
(D) [tex]\(25y^2 + 25x^3 = C\)[/tex] does not satisfy the equation.
(E) [tex]\(23y^2 + 2x^2 = C\)[/tex] does not satisfy the equation.
(F) [tex]\(2y^3 + 25x^3 = C\)[/tex] satisfies the equation with [tex]\(C = -\frac{1}{12}\)[/tex].
(G)[tex]\(23y^2 + 23x^2 = C\)[/tex] satisfies the equation with [tex]\(C = -\frac{1}{12}\)[/tex].
(H) [tex]\(23y^3 + 25x^3 = C\)[/tex] does not satisfy the equation.
Therefore, the orthogonal trajectories are given by options (C), (F), and (G), i.e., [tex]\(y^2 + 2x^2 = C\)[/tex],
[tex]\(2y^3 + 25x^3 = C\)[/tex], and
[tex]\(23y^2 + 23x^2 = C\)[/tex].
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