Question

Let \( F \) be the family of curves given by \[ x^2 + 2hxy + y^2 = 1, \quad -1<h<1. \] Then, the differential equation for the family of orthogonal trajectories to \( F \) is:

• A. \( (x^2 y - y^3 + y) \frac{dy}{dx} - (xy^2 - x^3 + x) = 0 \)
• B. \( (x^2 y - y^3 + y) \frac{dy}{dx} + (xy^2 - x^3 + x) = 0 \)
• C. \( (x^2 y + y^3 + y) \frac{dy}{dx} - (xy^2 + x^3 + x) = 0 \)
• D. \( (x^2 y + y^3 + y) \frac{dy}{dx} + (xy^2 + x^3 + x) = 0 \)

Hint:

When finding the differential equation of orthogonal trajectories, use implicit differentiation and then take the negative reciprocal of the slope.

Answer 1

The Correct Option is A

The given family of curves is: \[ x^2 + 2hxy + y^2 = 1. \] To find the equation of the orthogonal trajectories, we first differentiate the equation implicitly with respect to \( x \): \[ \frac{d}{dx}\left(x^2 + 2hxy + y^2\right) = 0. \] Using the product rule and the chain rule: \[ 2x + 2h \left( x \frac{dy}{dx} + y \right) + 2y \frac{dy}{dx} = 0. \] This simplifies to: \[ 2x + 2h y + 2h x \frac{dy}{dx} + 2y \frac{dy}{dx} = 0. \] Rearranging to isolate \( \frac{dy}{dx} \): \[ \left( 2h x + 2y \right) \frac{dy}{dx} = -2x - 2h y. \] Thus, the slope of the tangent to the family of curves is given by: \[ \frac{dy}{dx} = \frac{-2x - 2h y}{2h x + 2y}. \] The slope of the orthogonal trajectories will be the negative reciprocal of this: \[ \frac{dy}{dx} = \frac{2h x + 2y}{2x + 2h y}. \] Multiplying both sides by \( (x^2 y - y^3 + y) \), we get the differential equation for the orthogonal trajectories: \[ (x^2 y - y^3 + y) \frac{dy}{dx} - (xy^2 - x^3 + x) = 0. \] Thus, the correct answer is (A).