Question

What is the envelope of straight lines given by x cos b + y sin b = a sec b, where b is the parameter?

(a) y^2 + 4a(a-x) = 0

(b) y + 4ax = 0

(c) y + 4a(a-x) = 0

(d) y^2 = 4a(a-x)

I had been asked this question by my school teacher while I was bunking the class.

My question is from Envelopes topic in division Differential Calculus of Engineering Mathematics

Answer 1

Correct choice is (d) y^2 = 4a(a-x)

For explanation I would say: Dividing the given equation by cos b,

x + y tan b = a \(\frac{sec⁡b}{cos⁡b}\) = a sec^2 b = a(1 + tan^2 b)

The above equation can be written as   a tan^2 b – y tan b + (a-x) = 0 which is a quadratic equation in tan b

Hence, A = a, B = -y, C = a-x

The envelope is B^2 – 4AC = 0

(-y)^2 – 4(a)(a-x) = 0

y^2 = 4a(a-x).