Q45 / 160 · TS EAMCET 2014 · Mathematics
If $x^2+\alpha y^2+2 \beta y=a^2$ represents a pair of perpendicular lines, then $\beta$ equals to
- A$4 a$
- Correct answerB$a$
- C$2a$
- D$3a$
Solution
Let given line be
$$
x^2+\alpha y^2+2 \beta y-a^2=0
$$
Here, $a=1, b=\alpha, h=0, g=0, f=\beta, c=-a^2$
Condition for perpendicular line $a+b=0$
$$
\therefore \quad 1+\alpha=0 \Rightarrow \alpha=-1
$$
Condition of pair of lines
$$
\begin{array}{ccc}
& a b c+2 f g h-a f^2-b g^2-c h^2=0 \\
\therefore & 1 \times \alpha \times\left(-a^2\right)+0-1(\beta)^2-0-\left(-a^2\right)(0)=0 \\
\Rightarrow & -a^2 \alpha-\beta^2=0 \\
\Rightarrow & \beta^2=-\alpha a^2 \\
\Rightarrow & \beta^2=-(-1) a^2 \quad(\because \alpha=-1) \\
\Rightarrow & \beta^2=a^2 \Rightarrow \beta=a
\end{array}
$$
Hence, the correct answer is option: 'B'.