The asymptotes of the hyperbola `(x^(2))/(a_(1)^(2))-(y^(2))/(b_(1)^(2))=1` and `(x^(2))/(a_(2)^(2))-(y^(2))/(b_(2)^(2))=1` are perpendicular to each other. Then, (a) `a_(1)/a_(2)=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)`

Options

A. `a_(1)//a_(2)=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

C

Explanation

The slopes of asymptotes are
`m_(1)=(b_(1))/(a_(1)),m_(2)=(b_(2))/(a_(2))`
According to the question,
`m_(1)m_(2)=-1`
`"or "a_(1)a_(2)+b_(1)b_(2)=0`