Let $a$ and $b$ be non-zero real numbers. Then, the equation $(ax^{2} + by^{2} + c)(x^{2} - 5xy + 6y^{2}) = 0$ represents

$(ax^2+by^2+c)(x^2-5xy+6y^2)=0$

$a,\ b \neq 0$

$\text{Each factor } = 0$

$\text{To find: what the combined equation represents}$

$x^2-5xy+6y^2=x^2-2xy-3xy+6y^2$

$=(x-2y)(x-3y)$

$x-2y=0 \Rightarrow x=2y$

$x-3y=0 \Rightarrow x=3y$

$\text{Product}=0 \Rightarrow \text{union of the two loci}$

$\text{Plan: classify } ax^2+by^2+c=0 \text{ by signs of } a,b,c$

$\text{then combine with the two lines}$

$ax^2+by^2=-c$

$a=b:\ \text{circle};\quad a,b \text{ same sign}:\ \text{ellipse}$

$a,b \text{ opposite sign}:\ \text{hyperbola}$

$(1):\ c=0,\ a,b \text{ same sign} \Rightarrow ax^2+by^2=0$

$\Rightarrow \text{only } (0,0) \text{ (a point)}$

$c=0,\ a,b \text{ opposite} \Rightarrow \text{two lines} \Rightarrow \text{total } 4 \text{ lines}$

$(2):\ a=b \Rightarrow a(x^2+y^2)+c=0$

$x^2+y^2=-\dfrac{c}{a}$

$c \text{ opposite to } a \Rightarrow -\dfrac{c}{a}>0$

$\Rightarrow \text{real circle, radius } \sqrt{-c/a}$

$(3):\ a,b \text{ same},\ c \text{ opposite} \Rightarrow \dfrac{x^2}{-c/a}+\dfrac{y^2}{-c/b}=1$

$\Rightarrow \text{ellipse, not hyperbola}$

$(4):\ \text{two lines} + \text{one ellipse} \neq \text{circle and ellipse}$

$\text{Correct: } (2)\ \text{two straight lines and a circle}$