Question

If $(a{x}^2+bx+c)y+a^{\prime }{x}^2+b^{\prime }x+c^{\prime }=0$, then the condition that x may be rational function of $y$ is

• A. $(a{b}^{\prime }-a^{\prime }b{)}^2=(a{c}^{\prime }-a^{\prime }c\left)\right(b{c}^{\prime }-b^{\prime }c)$
• B. $(b{c}^{\prime }-b^{\prime }c{)}^2=(a{b}^{\prime }-a^{\prime }b\left)\right(a{c}^{\prime }-a^{\prime }c)$
• C. $(a{c}^{\prime }-a^{\prime }c{)}^2=(a{b}^{\prime }-a^{\prime }b\left)\right(b{c}^{\prime }-b^{\prime }c)$
• D. None of the above

Answer 1

The correct option is C $(a{c}^{\prime }-a^{\prime }c{)}^2=(a{b}^{\prime }-a^{\prime }b\left)\right(b{c}^{\prime }-b^{\prime }c)$

The given equation can be written as
$(ay+a^{\prime })x^2+(by+b^{\prime })x+(cy+c^{\prime })=\mathrm{0...}\left(i\right)$
The condition that $x$ may be rational function of $y$ is that the discriminant of $\left(i\right)$ should be a perfect square,
$(by+b^{\prime }{)}^2-4(ay+a^{\prime }\left)\right(cy+c^{\prime })$ is a perfect square.
$\Rightarrow (b^2-4ac)y^2+(2b{b}^{\prime }-4a{c}^{\prime }-4a^{\prime }c)y+b^{\prime 2}-4a^{\prime }{c}^{\prime }$ is a perfect square.
Which is true if $D=0$,
$\Rightarrow 4(b{b}^{\prime }-2a{c}^{\prime }–2a^{\prime }c{)}^2-4(b^2-4ac\left)\right(b^{\prime 2}-4a^{\prime }{c}^{\prime })=0$
$\Rightarrow (a{c}^{\prime }+a^{\prime }c)-4a{a}^{\prime }c{c}^{\prime }=ab{b}^{\prime }c+a^{\prime }b{b}^{\prime }c–a^{\prime }b{b}^{\prime }c–a^{\prime }{c}^{\prime }{b}^2-ac{b}^{\prime 2}$
$\Rightarrow (a{c}^{\prime }-a^{\prime }c{)}^2=(a{b}^{\prime }-a^{\prime }b\left)\right(b{c}^{\prime }-b^{\prime }c)$