Find the condition that one root of $ax^2+bx+c=0$ be the reciprocal of a root of $a_1x^2+b_1x+c_1=0$. Find the condition that one root of $ax^2+bx+c=0$ be the reciprocal of a root of $a_1x^2+b_1x+c_1=0$. 

My Attempt:

Let $x_1$ and $x_2$ be the roots of the equation $ax^2+bx+c=0$. Then,
$$x_1=\dfrac {-b+\sqrt {b^2-4ac}}{2a}$$
 $$x_2=\dfrac {-b-\sqrt {b^2-4ac}}{2a}$$
Again, Let $x'_1$ and $x'_2$ be the roots of $a_1x^2+b_1x+c_1=0$. Then,
$$x'_1=\dfrac {-b_1+\sqrt {{b_1}^2-4a_1c_1}}{2a_1}$$
$$x'_2=\dfrac {-b_2+\sqrt {{b_2}^2-4a_2c_2}}{2a_2}$$

How do I proceed further?
In general if $\alpha$ and $\beta$ are the roots of equation $ax^2+bx+c=0$, then,

Sum of the roots, $\alpha+\beta=-\dfrac{b}{a}$, and Multiplication of roots, $\alpha\beta=\dfrac{c}{a}$.

So, $\dfrac{\alpha+\beta}{\alpha\beta}=\dfrac{-b/a}{c/a}=\dfrac{-b}{c}\implies\dfrac{1}{\alpha}+\dfrac{1}{\beta}=\dfrac{-b}{c}$, and $\dfrac{1}{\alpha\beta}=\dfrac{a}{c}\ \hspace{20pt}\cdots(\text{i})$

Let, $\alpha_1$ and $\beta_1$ are the roots of equation $a_1x^2+b_1x+c_1=0$  such that $\alpha_1=\dfrac{1}{\alpha}$ and $\beta_1=\dfrac{1}{\beta}$, as they are reciprocal.

Now from (i) we can write that $\alpha_1+\beta_1=-\dfrac{b}{c}$ and $\alpha_1\beta_1=\dfrac{a}{c}$.

So the equation whose sum of roots is $=-\dfrac{b}{c}$ and multiplication of roots is $=\dfrac{a}{c}$ can be written as: $x^2-(\alpha_1+\beta_1)x+\alpha_1\beta_1=0\implies x^2+\dfrac{b}{c}x+\dfrac{a}{c}=0\implies cx^2+bx+a=0$. Now this final equation $cx^2+bx+a=0$ and $a_1x^2+b_1x+c_1=0$ are same. 

So comparing we get, $\dfrac{a_1}{c}=\dfrac{b_1}{b}=\dfrac{c_1}{a}$.
In general if $\alpha$ and $\beta$ are the roots of equation $ax^2+bx+c=0$, then,

Sum of the roots, $\alpha+\beta=-\dfrac{b}{a}$, and Multiplication of roots, $\alpha\beta=\dfrac{c}{a}$.

So, $\dfrac{\alpha+\beta}{\alpha\beta}=\dfrac{-b/a}{c/a}=\dfrac{-b}{c}\implies\dfrac{1}{\alpha}+\dfrac{1}{\beta}=\dfrac{-b}{c}$, and $\dfrac{1}{\alpha\beta}=\dfrac{a}{c}\ \hspace{20pt}\cdots(\text{i})$

Let, $\alpha_1$ and $\beta_1$ are the roots of equation $a_1x^2+b_1x+c_1=0$  such that $\alpha_1=\dfrac{1}{\alpha}$ and $\beta_1=\dfrac{1}{\beta}$, as they are reciprocal.

Now from (i) we can write that $\alpha_1+\beta_1=-\dfrac{b}{c}$ and $\alpha_1\beta_1=\dfrac{a}{c}$.

So the equation whose sum of roots is $=-\dfrac{b}{c}$ and multiplication of roots is $=\dfrac{a}{c}$ can be written as: $x^2-(\alpha_1+\beta_1)x+\alpha_1\beta_1=0\implies x^2+\dfrac{b}{c}x+\dfrac{a}{c}=0\implies cx^2+bx+a=0$. Now this final equation $cx^2+bx+a=0$ and $a_1x^2+b_1x+c_1=0$ are same. 

So comparing we get, $\dfrac{a_1}{c}=\dfrac{b_1}{b}=\dfrac{c_1}{a}$.