The condition that the straight-line `lx + my = n` may be a normal to the hyperbola `b^2x^2-a^2y^2 =a^2b^2` is:

Options

A. `(a^(2))/(l^(2))+(b^(2))/(m^(2))=((a^(2)+b^(2))^(2))/(n^(2))`
B. `(a^(2))/(l^(2))-(b^(2))/(m^(2))=((a^(2)-b^(2))^(2))/(n^(2))`
C. `(a^(2))/(l^(2))-(b^(2))/(m^(2))=((a^(2)+b^(2))^(2))/(n^(2))`
D. none of these

Explanation

The equation of the normal at `(a sec phi,b tan phi)` to the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` is
`axsinphi+by=(a^(2)+b^(2)) tan phi` ……….`(i)`
and the equation of the line is
`lx+my+n=0`…….`(ii)`
Now `(i)`, and `(ii)` represent the same line.
`:.(asinphi)/(1)=(b)/(m)=((a^(2)+b^(2))tanphi)/(-n)=((a^(2)+b^(2))sinphi)/(-ncosphi)`
`impliessinphi=(bl)/(am)` and `cosphi=((a^(2)+b^(2))l)/(-na)`
`implies1=(b^(2)t^(2))/(a^(2)m^(2))+((a^(2)+b^(2))^(2)l^(2))/(n^(2)a^(2))` [Squaring and adding]
`implies(a^(2))/(l^(2))-(b^(2))/(m^(2))=((a^(2)+b^(2))^(2))/(n^(2))`