Question

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

Answer 1

Given straight line $\frac{x}{\alpha }+\frac{y}{\beta }=1$ or $\frac{y}{\beta }=1-\frac{x}{\alpha }$

Squaring both sides, we get

$\frac{y^2}{{\beta }^2}=1+\frac{x^2}{{\alpha }^2}-\frac{2x}{\alpha }\Rightarrow \frac{{\alpha }^2}{{\beta }^2}=\frac{{\alpha }^2-2x\alpha +x^2}{y^2}$ ....(1)

Also, $\frac{b^2}{{\alpha }^2}+\frac{a^2}{{\beta }^2}=1\Rightarrow \frac{{\alpha }^2}{{\beta }^2}=\frac{{\alpha }^2-b^2}{a^2}$ ...(2)

Eliminating $\frac{{\alpha }^2}{{\beta }^2}$ from (1) using (2), we get

$\frac{{\alpha }^2-b^2}{a^2}=\frac{{\alpha }^2-2x\alpha +x^2}{y^2}$

$\to y^2{\alpha }^2-b^2{y}^2=a^2{\alpha }^2-2a^{2}x\alpha +a^2{x}^2$

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

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

Therefore, $4a^4{x}^2+4(y^2-a^2)(a^2{x}^2+b^2{y}^2)=0$

$\Rightarrow a^4{x}^2+a^2{x}^2{y}^2+b^2{y}^4-a^4{x}^2-a^2{b}^2{y}^2=0$

$\Rightarrow a^2{x}^2+b^2{y}^2=a^2{b}^2$

$\Rightarrow \frac{x^2}{b^2}+\frac{y^2}{a^2}=1$