Question

Find the envelope of the straight line $\frac{x}{\alpha }+\frac{y}{\beta }=1$, when $a\alpha +b\beta =c$.

Answer 1

Given straight line $\frac{x}{\alpha }+\frac{y}{\beta }=1$ ....(1)

Also, $a\alpha +b\beta =c⟹\beta =\frac{c-a\alpha }{b}$ ....(2)

Eliminating $\beta$ from (1) using (2), we get

$\frac{x}{\alpha }+\frac{by}{c-a\alpha }=1$

$\Rightarrow cx-cx\alpha +by\alpha =c\alpha -a{\alpha }^2$

$\Rightarrow a{\alpha }^2+(by-ax-c)\alpha +cx=0$

This is a quadratic equation and to find the envelop we need to equate the discriminant to zero.

Therefore, $(by-ax-c{)}^2=4acx$