Question

The asymptotes of the hyperbola $\frac{x^2}{a_1^2}-\frac{y^2}{b_1^2}=1$ and $\frac{x^2}{a_2^2}-\frac{y^2}{b_2^2}=1$ are perpendicular to each other, then :

• A. $\frac{a_1}{a_2}=\frac{b_1}{b_2}$
• B. $a_1{a}_2=b_1{b}_2$
• C. $a_1{a}_2+b_1{b}_2=0$
• D. $a_1-a_2=b_1-b_2$

Answer 1

Answer: (B), (C)

$a_1{a}_2=b_1{b}_2$
$a_1{a}_2+b_1{b}_2=0$

Asymptotes of the hyperbola $\frac{x^2}{a_1^2}-\frac{y^2}{b_1^2}=1$ is
$y=\pm \frac{b_1}{a_1}x$
Asymptotes of the hyperbola $\frac{x^2}{a_2^2}-\frac{y^2}{b_2^2}=1$ is
$y=\pm \frac{b_2}{a_2}x$
Slopes of asymptotes are
$m_1=\pm \frac{b_1}{a_1},m_2=\pm \frac{b_2}{a_2}$
According to the question,
$m_1{m}_2=-1$ (taking $m_1,m_2$ both of same sign)
$\Rightarrow a_1{a}_2+b_1{b}_2=0$
Now, taking $m_1,m_2$ of opposite sign, we get
$\Rightarrow a_1{a}_2=b_1{b}_2$