Question

Find the envelope of the family of straight line x/a+y/b=1 where a and b connected by the relation is ab=c²

Answer 1

The envelope of the family of straight lines x/a + y/b = 1 is 4xy = c²

Step-by-step explanation:

The given straight lines are

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

and the given relation is

ab = c² ..... (2)

Taking a as independent variable and from (1) and (2), differentiating both sides with respect to a, we get

x/a² + y/b² (db/da) = 0 ..... (3)

1/a + 1/b (db/da) = 0 .... (4)

From (3) and (4), we get

(x/a²)/(1/a) = (y/b²)/(1/b)

or, x/a = y/b = (x/a + y/b)/(1 + 1) = 1/2 [by (1)]

Then x/a = 1/2 or, a = 2x

and y/b = 1/2 or, b = 2y

From (2), putting values of a, b in terms of x, y, we have

(2x) (2y) = c²

or, 4xy = c²

This is the equation of the required envelope.

Answer 2

Answer:

Heya!

Step-by-step explanation:

Please refer to the attachment!!