Solve y = px − √(m²+p²)

y = px − √(m² + p²) is Clairaut's equation, y = px + f(p), with f(p) = −√(m² + p²).

General solution

Differentiate with respect to x (p = dy/dx):

p = p + x dp/dx − [p/√(m² + p²)] dp/dx

(dp/dx)[x − p/√(m² + p²)] = 0

The factor dp/dx = 0 gives p = c (constant). Substituting in the equation:

y = cx − √(m² + c²) (general solution, a family of straight lines)

Singular solution

The other factor gives x = p/√(m² + p²). Hence

x²(m² + p²) = p² → p²(1 − x²) = m²x² → p = mx/√(1 − x²) (taking this sign)

Then √(m² + p²) = m/√(1 − x²), and

y = px − √(m² + p²) = m x²/√(1 − x²) − m/√(1 − x²) = −m√(1 − x²)

Squaring: y² = m²(1 − x²), that is

m²x² + y² = m² (singular solution)

This ellipse x² + y²/m² = 1 is the envelope of the family of lines. (It is also found by eliminating c between y = cx − √(m² + c²) and 0 = x − c/√(m² + c²).)