For \(x\) to be rational function of \(y\), the term \((by+b')^2-4(ay+a')(cy+c')\) should be perfect square, giving square root as a linear in \(y\) , or it could be zero.

Thus I think, conditions will be as following -

\((i) \quad (b^2-4ac)y^2-2(2ac'+2a'c-bb')y+(b'^2-4a'c')=0\) ... i.e. giving 2 values of \(y\)

\((ii)\quad\) Comparing the above quadratic with \(kx^2+lx+m=0\), condition for perfect square is \(\Bigl( \dfrac{l}{2k} \Bigr)^2 = \dfrac{m}{k} \implies l^2=4mk \implies (2ac'+2a'c-bb')^2 = (b^2-4ac)(b'^2-4a'c') \)

@Rajat Bisht if you have something like answer key to check. please tell me if this is right.

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## Comments

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TopNewestHere's one of the things that I could instantaneously think of, maybe it's one way...

\((ay+a')x^2+(by+b')x+(cy+c') =0 \)

Solving the quadratic, its roots will be

\(x=\dfrac{-by-b' \pm \sqrt{(by+b')^2-4(ay+a')(cy+c')}}{2ay+2a'}\)

For \(x\) to be

rationalfunction of \(y\), the term \((by+b')^2-4(ay+a')(cy+c')\) should be perfect square, giving square root as a linear in \(y\) , or it could be zero.Thus I think, conditions will be as following -

\((i) \quad (b^2-4ac)y^2-2(2ac'+2a'c-bb')y+(b'^2-4a'c')=0\) ... i.e. giving 2 values of \(y\)

\((ii)\quad\) Comparing the above quadratic with \(kx^2+lx+m=0\), condition for perfect square is \(\Bigl( \dfrac{l}{2k} \Bigr)^2 = \dfrac{m}{k} \implies l^2=4mk \implies (2ac'+2a'c-bb')^2 = (b^2-4ac)(b'^2-4a'c') \)

@Rajat Bisht if you have something like answer key to check. please tell me if this is right.

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Bingo you got it right aditya..thanks well for ur convenience the problem is from hall &knight

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@megh choksi @Pratik Shastri @brian charlesworth @Aditya Raut @Azhaghu Roopesh M @Pranjal Jain

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