Question

Find the envelope of the family of straight lines x/a+y/b=1,where the parameter a and b are connected by a2+b2=c2, c being a fixed constant

Answer 1

The required envelope is

x^(2/3) + y^(2/3) = c^(2/3)

Step-by-step explanation:

The given family of straight lines is

x/a + y/b = 1 ..... (1)

and the relation between the parameters is

a² + b² = c² ..... (2)

Differentiating both sides of (1) and (2) by a, we get

- x/a² - y/b² db/da = 0 or, db/da = - (x/a²)/(y/b²) &

2a + 2b db/da = 0 or, db/da = - a/b

Equating db/da, we get

- (x/a²)/(y/b²) = - a/b

or, (x/a²)/a = (y/b²)/b

or, (x/a)/a² = (y/b)/b²

or, (x/a)/a² = (y/b)/b² = (x/a + y/b)/(a² + b²) = 1/c²

or, x/a³ = y/b³ = 1/c²

or, a³/x = b³/y = c²

∴ a³ = c²x, b³ = c²y

From (2), we get

(a³)^(2/3) + (b³)^(2/3) = c²

or, (c²x)^(2/3) + (c²y)^(2/3) = c²

or, c^(4/3) * {x^(2/3) + y^(2/3)} = c²

or, x^(2/3) + y^(2/3) = c^(2 - 4/3)

or, x^(2/3) + y^(2/3) = c^(2/3)

This is the equation of the required envelope.