Question

Find the envelope of the straight line $\frac{x}{\alpha }+\frac{y}{\beta }=1$, when $\alpha +\beta +\sqrt{{\alpha }^2+{\beta }^2}=c$.

Answer 1

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

Also, $\alpha +\beta +\sqrt{{\alpha }^2+{\beta }^2}=c$ ....(2)

Solving equation (2) further, we get

$\sqrt{{\alpha }^2+{\beta }^2}=c-\alpha -\beta$

$\Rightarrow {\alpha }^2+{\beta }^2={\alpha }^2+{\beta }^2+c^2-2c\alpha -2c\beta +2\alpha \beta$

$\Rightarrow \beta =\frac{c(c-2\alpha }{2(c-\alpha )}$ ....(3)

Eliminating $\beta$ from (1) using (3), we get

$\frac{x}{\alpha }+\frac{2(c-\alpha )y}{c(c-2\alpha )}=1$

$\Rightarrow c^{2}x-2cx\alpha +2cy\alpha -2y{\alpha }^2=c^2\alpha -2c{\alpha }^2$

$\Rightarrow 2(y-c){\alpha }^2+c(c+2x-2y)\alpha -c^{2}x=0$

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

Therefore, $c^2(c+2x-2y{)}^2-8(y-c)c^{2}x=0$

$\Rightarrow c^2+4x^2+4y^2+4cx-4cy-8xy-8xy+8cx=0$

$\Rightarrow x^2+y^2-c(x+y)+\frac{c^2}{4}=0$