Differential Equations
Differential Equations
star_batch_jee_advanced_2025
Grade 12
Question:
The orthogonal trajectories of the family of coaxial circles $x^2 + y^2 + 2gx + C = 0$, where $g$ is a parameter are
family of circles with centre on y-axis
system of coaxial parabolas
x^2 + y^2 - Cx - Cy = 0, where C' is an arbitrary constant
system of circles with centre on x-axis
Step-by-Step Solution
Key Concept: Orthogonal trajectories are found by replacing $\frac{dy}{dx}$ with $-\frac{dx}{dy}$ in the differential equation of the original family.
Given $x^2 + y^2 + 2gx + C = 0$, differentiate to get $2x + 2y\frac{dy}{dx} + 2g = 0$, yielding $g = -\left(x + y\frac{dy}{dx}\right)$. For orthogonal trajectories, replace $\frac{dy}{dx}$ with $-\frac{dx}{dy}$: $y^2 - x^2 - 2xy\frac{dx}{dy} + C = 0$. Rearranging gives $2x\frac{dx}{dy} - \frac{1}{y}x^2 = \frac{-C}{y} - y$. Substituting $x^2 = v$ produces the linear equation $\frac{dv}{dy} - \frac{1}{y}v = \frac{-C}{y} - y$. Solving with integrating factor $\frac{1}{y}$ yields $x^2 + y^2 - ky - C = 0$, which represents circles centered on the $y$-axis.
Correct Answer: 1