Let $a$ and $b$ be non-zero and real numbers. Then, the equation $\left(a x^2+b y^2+c\right)\left(x^2-5 x…
Let $a$ and $b$ be non-zero and real numbers. Then, the equation $\left(a x^2+b y^2+c\right)\left(x^2-5 x y+6 y^2\right)=0$ represents
Four straight lines, when $c=0$ and $a, b$ are of the same sign
Two straight lines and a circle, when $a=b$ and $c$ is of sign opposite to that of $a$
Two straight lines and a hyperbola, when $a$ and $b$ are of the same sign and $c$ is of sign opposite to that of $a$
A circle and an ellipse, when $a$ and $b$ are of the same sign and $c$ is of sign opposite to that of $a$
Solution
Let $a$ and $b$ be non- zero real numbers.
Therefore, the given equation $\left(a x^2+b y^2+c\right)\left(x^2-5 x y+6 y^2\right)=0$ implies either
$
\begin{array}{rc}
& x^2-5 x y+6 y^2=0 \\
\Rightarrow & (x-2 y)(x-3 y)=0 \\
\Rightarrow & x=2 y \text { and } x=3 y
\end{array}
$
represent two straight lines passing through origin.
Or $\quad a x^2+b y^2+c=0$
When $c=0$ and $a$ and $b$ are of same signs, then
$
\begin{aligned}
& a x^2+b y^2+c=0 \\
\Rightarrow & x=0 \text { and } y=0
\end{aligned}
$
which is a point specified as the origin.
When $a=b$ and $c$ is of sign opposite to that of $a, a x^2+b y^2+c=0$ represents a circle.
Hence, the given equation $\left(a x^2+b y^2+c\right)\left(x^2-5 x y+6 y^2\right)=0$
may represent two straight lines and a circle