Question

Let \( C \) denote the family of curves described by \( yx^2 = \lambda \), for \( \lambda \in (0, \infty) \) and lying in the first quadrant of the \( xy \)-plane. Let \( O \) denote the family of orthogonal trajectories of \( C \). Which one of the following curves is a member of \( O \), and passes through the point \( (2, 1) \)?

• A. \( y = \frac{x^2}{4}, \, x>0, y>0 \)
• B. \( x^2 - 2y^2 = 2, \, x>0, y>0 \)
• C. \( x - y = 1, \, x>0, y>0 \)
• D. \( 2x - y^2 = 3, \, x>0, y>0 \)

Hint:

To find orthogonal trajectories, differentiate the given family of curves, use the negative reciprocal of the slope for the orthogonal trajectory, and then solve the resulting differential equation.

Answer 1

The Correct Option is B

Step 1: Equation of Family \( C \).
The given family of curves is: \[ yx^2 = \lambda \] Differentiating implicitly with respect to \( x \): \[ \frac{d}{dx}(yx^2) = 0 \] Applying the product rule: \[ y' x^2 + 2xy = 0 \quad \Rightarrow \quad y' = -\frac{2y}{x} \] Step 2: Equation of Orthogonal Trajectories.
For orthogonal trajectories, the slope of the tangent lines satisfies: \[ y'_{\text{orth}} = \frac{x}{2y} \] Step 3: Solve for Orthogonal Trajectories.
The equation of the orthogonal trajectories is given by: \[ \frac{dy}{dx} = \frac{x}{2y} \] Separating variables: \[ 2y \, dy = x \, dx \] Integrating both sides: \[ y^2 = \frac{x^2}{2} + C \] Thus, the equation of the orthogonal trajectory is: \[ y^2 = \frac{x^2}{4} + C \] Using the point \( (2, 1) \), we find \( C = 0 \), so the equation of the orthogonal trajectory is: \[ y = \frac{x^2}{4} \] Final Answer: \[ \boxed{y = \frac{x^2}{4}, \, x>0, y>0} \]