Question

Find the envelope of the family of lines \(\frac{x}{t} \) + yt = 2c, t being the parameter.

(a) xy = c

(b) xy = 2c

(c) xy = 2

(d) xy = c^2

The question was asked during a job interview.

This interesting question is from Envelopes topic in portion Differential Calculus of Engineering Mathematics

Answer 1

The correct answer is (d) xy = c^2

For explanation: Given equation can be written as yt^2 – 2ct + x = 0 ——–> eq(1)

The envelope of At^2 + Bt + C = 0 is B^2 – 4AC = 0 ———> eq(2)

From eq(1), A = y, B= -2c, C = x

Putting the values in eq(2),

(-2c)^2 – 4(y)(x) = 0

4c^2 – 4xy = 0

xy = c^2.