Envelope of a family of lines

Calculus Level 3

For t > 0 t \gt 0 , define the family of lines in the x y xy plane, passing through ( t , 0 ) (t, 0) and ( 0 , 1 t ) \left (0, \dfrac{1}{t} \right ) . These lines define an envelope that is tangent to the lines where they meet. What is the equation of this envelope ?

Inspiration

y = 1 x y = \dfrac{1}{x} y = 1 4 x y = \dfrac{1}{4x} y = 1 − x y = 1 - x y = 1 2 x y = \dfrac{1}{2x}

Hosam Hajjir by Hosam Hajjir , 56, Canada

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1 solution

David Vreken
Feb 18, 2021

Each line in the family of lines has an equation of y = 0 − 1 t t − 0 x + 1 t y = \cfrac{0 -\frac{1}{t}}{t - 0}x + \cfrac{1}{t} or y = − 1 t 2 x + 1 t y = -\cfrac{1}{t^2}x + \cfrac{1}{t} .

The envelope is made up of maximum heights of these lines at a given x x -coordinate, so d y d t = 2 x t 3 − 1 t 2 = 0 \cfrac{dy}{dt} = \cfrac{2x}{t^3} - \cfrac{1}{t^2} = 0 , which solves to t = 2 x t = 2x .

Substituting t = 2 x t = 2x into y = − 1 t 2 x + 1 t y = -\cfrac{1}{t^2}x + \cfrac{1}{t} and simplifying gives y = 1 4 x \boxed{y = \cfrac{1}{4x}} .

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