Question

The orthogonal trajectories of the family of circles given by x² + y² – 2ay = 0 (a is parameter), is –

• A. x² + y² – 2kx = 0
• B. x² + y² – 2ky = 0
• C. x² + y² – 2k₁x – 2k₂y = 0
• D. None of these

Answer 1

Answer: (A)

x² + y² – 2kx = 0

Answer 2

The equation of the family of circles is x² + y² – 2ay

= 0... (1)

On differentiating w.r.t. x, we get

2x + 2y $\frac{dy}{dx}$ – 2a $\frac{dy}{dx}$ = 0 Ž a = $\frac{x + y\frac{dy}{dx}}{\frac{dy}{dx}}$

On putting this value of a in equation (1), we get

x² + y² – 2y $\left\{ \frac{x + y\frac{dy}{dx}}{\frac{dy}{dx}} \right\}$ = 0

Ž x² – y² – 2xy $\frac{dy}{dx}$ = 0.... (2)

This is the differential equation of the family of circles given by equation (1).

The differential equation representing the family of

orthogonal trajectories of equation (1) is obtained by replacing $\frac{dy}{dx}$ by $\left( –\frac{dx}{dy} \right)$ in equation (2). So, the differential equation of the orthogonal trajectories is

x² – y² + 2xy $\frac{dy}{dx}$ = 0

Ž (x² – y²) dx + 2xy dy = 0

Ž 2xy dy – y² dx = – x² dx

Ž$\frac{xd(y^{2}) - y^{2}dx}{x^{2}}$ = – dx

Ž d $\left( \frac{y^{2}}{x} \right)$ = – dx

Ž $\frac{y^{2}}{x}$ = – x + 2k. Ž x² + y² – 2kx = 0

This is the required family of orthogonal trajectories.

Hence (1) is the correct answer