Question

The pair of lines represented by $3a{x}^2+5xy+(a^2-2)y^2=0$ are perpendicular to each other for

• A. two values of $a$
• B. $a$
• C. one value of $a$
• D. no value of $a$

Answer 1

Answer: (A)

two values of $a$

For the lines to be perpendicular
coefficient of $x^2=-$coefficient of $y^2$
$3a=-(a^2-2)$
$a^2+3a-2=0$
$a=\frac{-3\pm \sqrt{9+8}}{2}$
$a=\frac{-3\pm \sqrt{17}}{2}$
Therefore, there are two values of $a.$