Let a and b be non-zero real numbers. Then,th | vedclass

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(B) The given equation is $(a x^2+b y^2+c)(x^2-5 x y+6 y^2)=0$. This implies $a x^2+b y^2+c=0$ or $x^2-5 x y+6 y^2=0$. The second part $x^2-5 x y+6 y^

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two straight lines and a circle,when $a=b$,and $c$ is of sign opposite to that of $a$

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(B) The given equation is $(a x^2+b y^2+c)(x^2-5 x y+6 y^2)=0$. This implies $a x^2+b y^2+c=0$ or $x^2-5 x y+6 y^2=0$. The second part $x^2-5 x y+6 y^2=0$ can be factored as $(x-2 y)(x-3 y)=0$,which represents two straight lines passing through the origin. For the first part $a x^2+b y^2+c=0$,if $a=b$ and $c$ has the opposite sign to $a$,then $x^2+y^2 = -c/a$,which represents a circle. Thus,the equation represents two straight lines and a circle when $a=b$ and $c$ is of sign opposite to that of $a$.

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

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