If the tangent and normal to a rectangular hyperbola cut off intercepts `x_(1)` and `x_(2)` on one axis and `y_(1)` and `y_(2)` on the other, then

Options

A. `a_(1)a_(2)+b_(1)b_(2)=0`
B. `a_(1)a_(2)=-b_(1)b_(2)`
C. `a_(1)b_(2)=a_(2)b_(1)`
D. `a_(1)a_(2)=b_(1)b_(2)`

Explanation

Let the equation of the rectangular hyperbola be
`x^(2)-y^(2)=a^(2)`
and, let `P(a sec theta, a tan theta)` be a point on it.
The equations of the tangent and normal at `P` are
`x sec theta-y tantheta=a`……..`(i)`
and, `x cos theta+y cot theta=2a`..........`(ii)`
It is given that the tangent at `P` cuts off intercepts `a_(1)` and `b_(1)` on the coordinate axes. Therefore,
`a_(1)=a cos theta` and `b_(1)=-a cot theta`
The normal at `P` cuts off intercepts `a_(2)` and `b_(2)` on the coordinate axes. Therefore,
`a_(2)=2a sec theta` and `b_(2)=2a tan theta`
Now,
`a_(1)a_(2)+b_(1)b_(2)=a cos thetaxx2a sec thet+(-a cot theta)(2a tantheta)`
`impliesa_(1)a_(2)+b_(1)b_(2)=2a^(2)-2a^(2)=0`