Question

Consider the family of curves \(x^2 - y^2 = ky\) with parameter \(k \in \mathbb{R}\). The equation of the orthogonal trajectory to this family passing through \((1,1)\) is given by

• A. \(x^3 + 3xy^2 = 4.\)
• B. \(x^2 + 2xy = 3.\)
• C. \(y^2 + 2x^2y = 3.\)
• D. \(x^3 + 2xy^2 = 3.\)

Hint:

Orthogonal trajectories are found by replacing \(y'\) with \(-1/y'\) in the differential equation of the given family, then solving the resulting first-order equation.

Answer 1

The Correct Option is A

Step 1: Differentiate the given family.
From \(x^2 - y^2 = ky \Rightarrow k = \frac{x^2 - y^2}{y}.\) Differentiate w.r.t. \(x\): \[ 2x - 2y y' = k y' + y k'. \] Eliminate \(k'\) and simplify to obtain the slope of the given family: \[ y' = \frac{2xy}{x^2 + y^2}. \]
Step 2: Orthogonal trajectory condition.
For the orthogonal trajectory, slope \(m_2 = -\frac{1}{y'} = -\frac{x^2 + y^2}{2xy}.\) Thus, \[ \frac{dy}{dx} = -\frac{x^2 + y^2}{2xy}. \]
Step 3: Separate and integrate.
Multiply both sides by \(2xy\): \[ 2xy\,dy = -(x^2 + y^2)\,dx. \] This is a homogeneous equation. Let \(y = vx \Rightarrow dy = v\,dx + x\,dv.\) Substitute and simplify to get \[ x^3(1 + 3v^2) = 4, \] which simplifies to \(x^3 + 3xy^2 = 4.\)
Step 4: Conclusion.
Hence, the orthogonal trajectory is \(\boxed{x^3 + 3xy^2 = 4}.\)