Question

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

• A. Two values of $a$
• B. For all $a$
• C. For one value of $a$
• D. For no value of $a$

Answer 1

The correct option is A

Two values of $a$

Explanation for the correct option

Given, $3a{x}^2+5xy+\left(a^2-2\right)y^2=0$

The condition for the lines to be perpendicular is

$$\begin{gathered} \mathrm{coefficient}\mathrm{of}{x}^2+\mathrm{coefficient}\mathrm{of}{y}^2=0 \\ \Rightarrow 3a+a^2-2=0 \\ \therefore a=\frac{-3\pm \sqrt{3^2-4\times 1\times \left(-2\right)}}{2} \\ a=\frac{-3\pm \sqrt{17}}{2} \end{gathered}$$

So, we get two values of $a$.

Hence, option $\left(\mathrm{A}\right)$ is the correct answer.