Question

Find the envelope of the family of straight line = − 2 − 3 , being the parameter.

Answer 1

axcosθ+bysinθ=c

2

…eqn(1)

Differentiating w.r.t x, we get

−axsinθ+bycosθ=0

⇒tanθ=

ax

by

Divide eqn (1) with cosθ, we get

ax+bytanθ=c

2

secθ

⇒ax+by(

ax

by

)=c

2

1+(

a

2

x

2

b

2

y

2

)

⇒a

2

x

2

+b

2

y

2

=c

4