Question

The equation $12x^2+7xy+a{y}^2+13a-y+3=0$ represents a pair of perpendicular lines. Then, the value of $a$ is

• A. $\frac{7}{2}$
• B. $-19$
• C. $-12$
• D. $12$

Answer 1

The correct option is C

$-12$

Explanation for the correct option:

Comparing $12x^2+7xy+a{y}^2+13a-y+3=0$ with the general equation of circle $A{x}^2+2hxy+B{y}^2+2gx+2fy+c=0$

We get $A=12,B=a$

Since the given equation represents pair of a perpendicular line

therefore, coefficient of $x^2$$+$coefficient of $y^2$$=0$

This gives

$$\begin{gathered} A+B=0 \\ \Rightarrow 12+a=0 \\ \Rightarrow a=-12 \end{gathered}$$

Hence, option (C) is the correct answer