The normal to the rectangular hyperbola `xy = 4` at the point `t_1` meets the curve again at the point `t_2` Then

Options

A. `t_(2)=-(1)/(t_(1)^(3))`
B. `t_(1)=-(1)/(t_(2)^(3))`
C. `t_(2)^(3)=-(1)/(t_(1)^(3))`
D. `t_(2)=(1)/(t_(1)^(3))`

Explanation

The equation of the normal at `(ct_(1),c//t_(1))` to the hyperbola `xy=c^(2)` is
`xt_(1)^(3)-yt_(1)-ct_(1)^(4)+c=0`
If this passes through `(ct_(2),(c )/(t_(2)))`, then
`ct_(2)t_(1)^(3)-(c )/(t_(2)t_(1)-ct_(1)^(4)+c=0`
`impliest_(2)^(2)t_(1)^(3)-t_(1)-t_(1)^(4)t_(2)+t_(2)=0`
`impliest_(1)^(3)t_(2)(t_(2)-t_(1))+(t_(2)-t_(1))=0`
`implies(t_(1)^(3)t_(2)+1)(t_(2)-t_(1))=0`
`impliest_(1)^(3)t_(2)+1=0` [` :' t_(2) ne t_(1)`]
`impliest_(2)=-1//t_(1)^(3)`