Orthogonal Trajectories
The family of curves y = Cx^2 has orthogonal trajectories that we must determine by replacing the slope with its negative reciprocal and solving the resulting equation.
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About This Question
- Subject
- mathematics
- Chapter
- differential equations
- Topic
- orthogonal trajectories
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
x2+2y2=K
Orthogonal trajectories are curves that intersect a given family at right angles, found by first eliminating the parameter to get the family's slope, then replacing that slope with its negative reciprocal. Starting from y = Cx^2, differentiate to get \frac{dy}{dx} = 2Cx. Eliminating C using C = \frac{y}{x^2} gives \frac{dy}{dx} = 2\cdot\frac{y}{x^2}\cdot x = \frac{2y}{x}, the slope of the original family. For orthogonal curves the slope becomes the negative reciprocal: \frac{dy}{dx} = -\frac{x}{2y}. This is separable: 2y,dy = -x,dx. Integrating gives y^2 = -\frac{x^2}{2} + \text{const}, which rearranges to x^2 + 2y^2 = K. Option x^2 - 2y^2 = K comes from a sign error in separation. Option 2x^2 + y^2 = K swaps the roles of x and y in the coefficient. Option x^2 + y^2 = K ignores the factor of 2 from the slope. This follows the standard JEE Advanced orthogonal-trajectory procedure. As a final geometric check, the result is a family of ellipses, which sensibly cross the parabolas y = Cx^2 transversally everywhere, consistent with perpendicular intersection.
This hard difficulty mathematics question is from the chapter differential equations, covering the topic of orthogonal trajectories. It appeared in the 2025 exam.
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