Question

if (x^3-1/y^3) proportional to (x^3+1/y^3) then prove x proportion to 1/y​

Answer 1

Answer:

Given:

[math]\frac{1}{y} -\frac{1}{x} \propto \frac{1}{x-y}\\\Rightarrow \frac{1}{y}-\frac{1}{x}=\frac{k}{x-y},\,k \text{ being the constant of variation.}\\\Rightarrow\frac{x-y}{xy}=\frac{k}{x-y}\\\Rightarrow (x-y)^2 = kxy\\\Rightarrow x^2 -2xy+y^2 =kxy\\\Rightarrow x^2 -(k+2)xy+y^2 =0\\\Rightarrow x^2 -mxy +y^2 =0,\text{ where }m=k+2\text{ is a}\\\text{constant.}\\\Rightarrow \left(\frac{x}{y}\right)^2 -m\left(\frac{x}{y}\right)+1=0[/math]

[math][\text{on dividing both the sides by }y^2.][/math]

[math]\Rightarrow \frac{x}{y} = \frac{m\pm \sqrt{m^2 -4}}{2}=l\,(\text{say}),\text{a constant.}[/math]

[math][\text{by quadratic formula.}][/math]

[math]\Rightarrow x =ly[/math]

This suggests that:

[math]\boxed{x \propto y.} \quad \dagger\tag*{}[/math]