- #1

rajeshmarndi

- 319

- 0

##ax^2+2hxy+by^2+2gx+2fy+c=0##

or ##ax2+2(hy+g)x+(by^2+2fy+c)=0 ## (quadratic in x)

##x = \frac{-2(hy+g)}{2a} ± \frac{√((hy+g)^2-a(by^2+2fy+c))}{2a}##

The terms inside square root need to be a perfect square and it is in quadratic in y i.e ## √[(h^2-ab)y^2+2(hg-af)y+(g^2-ac)]##

or ##√[Ay^2+By+C] = √[(√Ay+\frac{B}{2√A})^2-\frac{B^2-4AC}{2A}]##.

Hence the condition is taken as ## B^2-4AC=0 ##.

My question even if ## B^2-4AC≠0 ##, the term inside square root can still be a perfect square e.g ##4^2-7=3^2, 6^2-11=5^2,23^2-45=22^2## and so on...

Thank you.