Question

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

• A. Four straight lines, when c=0 and a, b are of the same sign
• B. Two straight lines and a circle, when a=b and c is of sign opposite to that of a
• C. 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
• D. A circle and an ellipse, when a and b are of the same sign and c is of sign opposite to that of a

Answer 1

The correct option is B Two straight lines and a circle, when a=b and c is of sign opposite to that of a

Let a and b non-zero real numbers.
Therefore the given equation
$(a{x}^2+b{y}^2+c)(x^2-5xy+6y^2)=0$ implies either
$x^2-5xy+6y^2=0$
$\Rightarrow (x-2y)(x-3y)=0$
$\Rightarrow x=2y$
And x=3y
Represent two straight lines passing through origin or $a{x}^2+b{y}^2+c=0$ when $c=0$ and a and b are of same signs, then
$a{x}^2+b{y}^2+c=0x=0$
And $y=0$
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,
$(a{x}^2+b{y}^2+c)(x^2-5xy+6y^2)=0$
May represent two straight lines and a circle.