Explanation

(b)
Given that $\alpha_{1},\alpha_{2}$ are the roots of the equation $ax^{2} + bx + c = 0$, then
$\alpha_{1} + \alpha_{2} = - \frac{b}{a}$ and $\alpha_{1}\alpha_{2} = \frac{c}{a}$ …(i)
Now, $\beta_{1},\beta_{2}$ are the roots of $px^{2} + qx + r = 0$, then
$\beta_{1} + \beta_{2} = - \frac{q}{p}$ and $\beta_{1}\beta_{2} = \frac{r}{p}$ …(ii)
Given system is $\alpha_{1}y + \alpha_{2}\mathcal{z} = 0$ and $\beta_{1}y + \beta_{2}\mathcal{z} = 0$.
$\Rightarrow \frac{\alpha_{1}}{\beta_{1}} = \frac{\alpha_{2}}{\beta_{2}}\$
Now, $\ \frac{\ \alpha_{1\alpha_{2}}}{\beta_{1}\beta_{2}} = \frac{\frac{c}{a}}{\frac{r}{p}}$
$\Rightarrow \ \frac{\alpha_{1}}{\beta_{1}}.\frac{\alpha_{2}}{\beta_{2}} = \frac{cp}{ar}$ …(iii)
Since, $\frac{\alpha_{1}}{\beta_{1}} = \frac{\alpha_{2}}{\beta_{2}} \Rightarrow \frac{\alpha_{1}}{\alpha_{2}} = \frac{\beta_{1}}{\beta_{2}} \Rightarrow \frac{\alpha_{1}^{2}}{\alpha_{2}^{2}} = \frac{\beta_{1}^{2}}{\beta_{2}^{2}}$
$\Rightarrow \ \frac{\alpha_{1}^{2} + \alpha_{2}^{2}}{\alpha_{2}^{2}} = \frac{\beta_{1}^{2} + \beta_{2}^{2}}{\beta_{2}^{2}}$ (on adding 1 on both sides)
$\Rightarrow \frac{\alpha_{2}^{2}}{\beta_{2}^{2}} = \frac{\alpha_{1}^{2} + \alpha_{2}^{2}}{\beta_{1}^{2} + \beta_{2}^{2}}$
$= \frac{\left( \alpha_{1} + \alpha_{2} \right)^{2} - 2\alpha_{1}\alpha_{2}}{\left( \beta_{1} + \beta_{2} \right)^{2} - 2\beta_{1}\beta_{2}}$
On substituting the values from Eqs. (i), (ii) and (iii), we get
$\frac{cp}{ar} = \frac{\frac{b^{2}}{a^{2}} - 2\left( \frac{c}{a} \right)}{\frac{q^{2}}{p^{2}} - 2\left( \frac{r}{p} \right)} = \frac{\left( b^{2} - 2ac \right)p^{2}}{\left( q^{2} - 2pr \right)a^{2}}$
$\Rightarrow \ \frac{c}{r} = \frac{pb^{2} - 2acp}{q^{2}a - 2apr}$
$\Rightarrow {\ b}^{2}rp - 2acpr = q^{2}ac - pr2ac$
$\Rightarrow b^{2}pr = q^{2}ac$